Q1 JEE Main 2026 · 8 APR · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
Consider the equation H = x p ϵ q E r t s H = \frac{x^p \epsilon^q E^r}{t^s} H = t s x p ϵ q E r
Where H = H = H = magnetic field intensity; E = E = E = electric field, ϵ = \epsilon = ϵ = permittivity, x = x = x = distance, t = t = t = time The values of p , q , r p, q, r p , q , r and s s s respectively are:
(a) 1 , 1 , 1 , 1 1, 1, 1, 1 1 , 1 , 1 , 1 (b) − 1 , 1 , 2 , 1 -1, 1, 2, 1 − 1 , 1 , 2 , 1 (c) 1 , − 1 , − 2 , 1 1,\ -1,\ -2, 1 1 , − 1 , − 2 , 1 (d) − 1 , − 2 , − 2 , 1 -1,\ -2,\ -2, 1 − 1 , − 2 , − 2 , 1 Show answer and solution
Answer: (a)
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Q2 JEE Main 2026 · 6 APR · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
The potential energy of a particle changes with distance x x x from a fixed origin as V = A x x + B V = \frac{A\sqrt{x}}{x+B} V = x + B A x , where A A A and B B B are constants with appropriate dimensions. The dimensions of A B AB A B are ______
(a) [ M 1 L 5 / 2 T − 2 ] \left[\mathrm{M^{1}L^{5/2}T^{-2}}\right] [ M 1 L 5/2 T − 2 ] (b) [ M 3 / 2 L 5 / 2 T − 2 ] \left[\mathrm{M^{3/2}L^{5/2}T^{-2}}\right] [ M 3/2 L 5/2 T − 2 ] (c) [ M 1 L 2 T − 2 ] \left[\mathrm{M^{1}L^{2}T^{-2}}\right] [ M 1 L 2 T − 2 ] (d) [ M 1 L 7 / 2 T − 2 ] \left[\mathrm{M^{1}L^{7/2}T^{-2}}\right] [ M 1 L 7/2 T − 2 ] Show answer and solution
Answer: (d)
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Q3 JEE Main 2026 · 5 APR · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
L , C L, C L , C and R R R represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula M L 2 T − 4 A − 2 \mathrm{ML^{2}T^{-4}A^{-2}} M L 2 T − 4 A − 2 corresponds to ______ .
(a) R L C \frac{R}{\sqrt{LC}} L C R (b) R L C \frac{R}{LC} L C R (c) C L R \frac{C}{\sqrt{LR}} L R C (d) 1 R L C \frac{1}{R}\sqrt{\frac{L}{C}} R 1 C L Show answer and solution
Answer: (a)
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Q4 JEE Main 2026 · 2 APR · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
The dimensional formula of 1 2 ϵ 0 E 2 \frac{1}{2}\epsilon_0 E^2 2 1 ϵ 0 E 2 (ϵ 0 \epsilon_0 ϵ 0 = permittivity of vacuum and E E E = electric field) is M a L b T c M^aL^bT^c M a L b T c . The value of 2 a − b + c = 2a - b + c = 2 a − b + c = ________.
Show answer and solution
Answer: (b)
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Q5 JEE Main 2026 · 21 JAN · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
Consider a modified Bernoulli equation.
( P + A B t 2 ) + ρ g ( h + B t ) + 1 2 ρ V 2 = \left(\mathrm{P} + \frac{A}{Bt^2}\right) + \rho g(h + Bt) + \frac{1}{2}\rho V^2 = ( P + B t 2 A ) + ρ g ( h + Bt ) + 2 1 ρ V 2 = constant
If t t t has the dimension of time then the dimensions of A A A and B B B are ______ respectively.
(a) [ M L 0 T − 1 ] \left[\mathrm{ML^{0}T^{-1}}\right] [ M L 0 T − 1 ] and [ M 0 L T ] [\mathrm{M^{0}LT}] [ M 0 LT ] (b) [ M L 0 T − 2 ] \left[\mathrm{ML^{0}T^{-2}}\right] [ M L 0 T − 2 ] and [ M 0 L T − 1 ] [\mathrm{M^{0}LT^{-1}}] [ M 0 L T − 1 ] (c) [ M L 0 T − 2 ] \left[\mathrm{ML^{0}T^{-2}}\right] [ M L 0 T − 2 ] and [ M 0 L T − 2 ] [\mathrm{M^{0}LT^{-2}}] [ M 0 L T − 2 ] (d) [ M L 0 T − 1 ] \left[\mathrm{ML^{0}T^{-1}}\right] [ M L 0 T − 1 ] and [ M 0 L T − 1 ] [\mathrm{M^{0}LT^{-1}}] [ M 0 L T − 1 ] Show answer and solution
Answer: (d)
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Q6 JEE Main 2026 · 2 APR · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
Dimensions of universal gravitational constant (G G G ) in terms of Planck's constant (h h h ), distance (L L L ), mass (M M M ) and time (T T T ) are _______.
(a) [ h T L M − 2 ] [hTLM^{-2}] [ h T L M − 2 ] (b) [ h T − 1 L M − 2 ] [hT^{-1}LM^{-2}] [ h T − 1 L M − 2 ] (c) [ h T L 2 M − 2 ] [hTL^2M^{-2}] [ h T L 2 M − 2 ] (d) [ h − 1 T − 1 L M − 2 ] [h^{-1}T^{-1}LM^{-2}] [ h − 1 T − 1 L M − 2 ] Show answer and solution
Answer: (b)
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Q7 NEET 2026 Easy ☆
Applications of Dimensional Analysis
The speed of light in vacuum is taken as unity. If light takes 6 min 40 s to reach the Earth from the Sun, the distance between the Sun and the Earth in new unit is:
(a) 3 × 1 0 8 3 \times 10^{8} 3 × 1 0 8 (b) 500 (c) 3 × 1 0 10 3 \times 10^{10} 3 × 1 0 10 (d) 400 Show answer and solution
Answer: (d)
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Q8 JEE Main 2026 · 8 APR · SHIFT 2 Medium ☆
Applications of Dimensional Analysis
A new unit ( α \alpha α ) of length is chosen such that it is equal to the speed of light in vacuum. What is the distance between Venus and Earth in terms of α \alpha α units if light takes 6 min. 40 s to cover this distance?
(a) 200 α 200\alpha 200 α (b) 400 α 400\alpha 400 α (c) 300 α 300\alpha 300 α (d) 500 α 500\alpha 500 α Show answer and solution
Answer: (b)
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Q9 JEE Main 2026 · 23 JAN · SHIFT 2 Medium ☆
Applications of Dimensional Analysis
A ball of radius r r r and density ρ \rho ρ dropped through a viscous liquid of density σ \sigma σ and viscosity η \eta η attains its terminal velocity at time t t t , given by t = A ρ a r b η c σ d t = A\rho^{a}r^{b}\eta^{c}\sigma^{d} t = A ρ a r b η c σ d , where A A A is a constant and a , b , c a, b, c a , b , c and d d d are integers. The value of b + c a + d \frac{b+c}{a+d} a + d b + c is ______ .
Numerical answer type. Work it out, then open the answer.
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Answer: 1
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Q10 JEE Main 2026 · 5 APR · SHIFT 2 Difficult ☆
Applications of Dimensional Analysis
Match List - I with List - II.
where h (Planck's constant), G (gravitational constant) and c (speed of light in vacuum) as
fundamental units.
Choose the correct answer from the options given below:
(a) A-II, B-IV, C-I, D-III (b) A-IV, B-II, C-I, D-III (c) A-IV, B-I, C-II, D-III (d) A-III, B-I, C-II, D-IV Show answer and solution
Answer: (b)
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Q11 JEE Main 2025 · 4 APR · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
In an electromagnetic system, the quantity representing the ratio of electric flux and magnetic flux has dimension of M P L Q T R A S M^PL^QT^RA^S M P L Q T R A S , where value of ' Q Q Q ' and ' R R R ' are
(a) ( 3 , − 5 ) (3,\ -5) ( 3 , − 5 ) (b) ( − 2 , 1 ) (-2, 1) ( − 2 , 1 ) (c) ( − 2 , 2 ) (-2, 2) ( − 2 , 2 ) (d) ( 1 , − 1 ) (1,\ -1) ( 1 , − 1 ) Show answer and solution
Answer: (d)
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Q12 JEE Main 2025 · 2 APR · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
The equation for real gas is given by ( P + a V 2 ) ( V − b ) = R T \left(\mathrm{P} + \frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V} - \mathrm{b}) = \mathrm{RT} ( P + V 2 a ) ( V − b ) = RT , where P , V , T \mathrm{P, V, T} P , V , T and R are the pressure, volume, temperature and gas constant, respectively. The dimension of a b − 2 \mathrm{ab^{-2}} a b − 2 is equivalent to that of:
(a) Compressibility (b) Planck's constant (c) Energy density (d) Strain Show answer and solution
Answer: (c)
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Q13 JEE Main 2025 · 29 JAN · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
The expression given below shows the variation of velocity (v) with time (t),
v = A t 2 + B t C + t v = \mathrm{At^{2}} + \frac{\mathrm{Bt}}{\mathrm{C+t}} v = A t 2 + C + t Bt .
The dimension of ABC is:
(a) [M0L1T−2] (b) [M0L2T−3] (c) [M0L2T−2] (d) [M0L1T−3] Show answer and solution
Answer: (b)
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Q14 JEE Main 2025 · 23 JAN · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
The position of a particle moving on x x x -axis is given by x ( t ) = A sin t + B cos 2 t + C t 2 + D x(t) = A\sin t + B\cos^2 t + Ct^2 + D x ( t ) = A sin t + B cos 2 t + C t 2 + D , where t t t is time. The dimension of A B C D \frac{ABC}{D} D A BC is
(a) L \mathrm{L} L (b) L 3 T − 2 \mathrm{L^{3}T^{-2}} L 3 T − 2 (c) L 2 T − 2 \mathrm{L^{2}T^{-2}} L 2 T − 2 (d) L 2 \mathrm{L^{2}} L 2 Show answer and solution
Answer: (c)
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Q15 NEET 2025 Medium ☆
Applications of Dimensional Analysis
A balloon is made of a material of surface tension S S S and its inflation outlet (from where gas is filled in it) has small area A A A . It is filled with a gas of density ρ \rho ρ and takes a spherical shape of radius R R R . When the gas is allowed to flow freely out of it, its radius r r r changes from R R R to 0 (zero) in time T T T . If the speed v ( r ) v(r) v ( r ) of gas coming out of the balloon depends on r r r as r a r^a r a and T ∝ S α A β ρ γ R δ T \propto S^\alpha A^\beta \rho^\gamma R^\delta T ∝ S α A β ρ γ R δ then
(a) a = − 1 2 , α = − 1 2 , β = − 1 , γ = 1 2 , δ = 7 2 a = -\frac{1}{2}, \alpha = -\frac{1}{2}, \beta = -1, \gamma = \frac{1}{2}, \delta = \frac{7}{2} a = − 2 1 , α = − 2 1 , β = − 1 , γ = 2 1 , δ = 2 7 (b) a = 1 2 , α = 1 2 , β = − 1 2 , γ = 1 2 , δ = 7 2 a = \frac{1}{2}, \alpha = \frac{1}{2}, \beta = -\frac{1}{2}, \gamma = \frac{1}{2}, \delta = \frac{7}{2} a = 2 1 , α = 2 1 , β = − 2 1 , γ = 2 1 , δ = 2 7 (c) a = 1 2 , α = 1 2 , β = − 1 , γ = + 1 , δ = 3 2 a = \frac{1}{2}, \alpha = \frac{1}{2}, \beta = -1, \gamma = +1, \delta = \frac{3}{2} a = 2 1 , α = 2 1 , β = − 1 , γ = + 1 , δ = 2 3 (d) a = − 1 2 , α = − 1 2 , β = − 1 , γ = − 1 2 , δ = 5 2 a = -\frac{1}{2}, \alpha = -\frac{1}{2}, \beta = -1, \gamma = -\frac{1}{2}, \delta = \frac{5}{2} a = − 2 1 , α = − 2 1 , β = − 1 , γ = − 2 1 , δ = 2 5 Show answer and solution
Answer: (a)
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Q16 NEET 2024 Easy ☆
Applications of Dimensional Analysis
The potential energy of a particle moving along x x x -direction varies as V = A x 2 x + B V = \frac{Ax^2}{\sqrt{x} + B} V = x + B A x 2 . The dimensions of A 2 B \frac{A^2}{B} B A 2 are:
(a) [ M 3 / 2 L 1 / 2 T − 3 ] \left[\mathrm{M^{3/2}L^{1/2}T^{-3}}\right] [ M 3/2 L 1/2 T − 3 ] (b) [ M 1 / 2 L T − 3 ] [\mathrm{M^{1/2}LT^{-3}}] [ M 1/2 L T − 3 ] (c) [ M 2 L 1 / 2 T − 4 ] [M^{2}L^{1/2}T^{-4}] [ M 2 L 1/2 T − 4 ] (d) [ M L 2 T − 4 ] \left[\mathrm{ML^{2}T^{-4}}\right] [ M L 2 T − 4 ] Show answer and solution
Answer: (c)
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Q17 NEET 2024 Easy ☆
Applications of Dimensional Analysis
A force defined by F = α t 2 + β t F = \alpha t^2 + \beta t F = α t 2 + βt acts on a particle at a given time t t t . The factor which is dimensionless, if α \alpha α and β \beta β are constants, is:
(a) β t / α \beta t/\alpha βt / α (b) α t / β \alpha t/\beta α t / β (c) α β t \alpha\beta t α βt (d) α β / t \alpha\beta/t α β / t Show answer and solution
Answer: (b)
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Q18 JEE Main 2024 · 9 APR · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
The de-Broglie wavelength associated with a particle of mass m m m and energy E E E is h / 2 m E h/\sqrt{2mE} h / 2 m E . The dimensional formula for Planck's constant is:
(a) [ M 2 L 2 T − 2 ] \left[\mathrm{M^{2}L^{2}T^{-2}}\right] [ M 2 L 2 T − 2 ] (b) [ M L − 1 T − 2 ] \left[\mathrm{ML^{-1}T^{-2}}\right] [ M L − 1 T − 2 ] (c) [ M L 2 T − 1 ] \left[\mathrm{ML^{2}T^{-1}}\right] [ M L 2 T − 1 ] (d) [ M L T − 2 ] [\mathrm{MLT^{-2}}] [ ML T − 2 ] Show answer and solution
Answer: (c)
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Q19 JEE Main 2024 · 5 APR · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
What is the dimensional formula of a b − 1 ab^{-1} a b − 1 in the equation ( P + a V 2 ) ( V − b ) = R T \left(\mathrm{P} + \frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V} - \mathrm{b}) = \mathrm{RT} ( P + V 2 a ) ( V − b ) = RT , where letters have their usual meaning.
(a) [ M 6 L 7 T 4 ] [\mathrm{M^{6}L^{7}T^{4}}] [ M 6 L 7 T 4 ] (b) [ M − 1 L 5 T 3 ] [\mathrm{M^{-1}L^{5}T^{3}}] [ M − 1 L 5 T 3 ] (c) [ M 0 L 3 T − 2 ] [\mathrm{M^{0}L^{3}T^{-2}}] [ M 0 L 3 T − 2 ] (d) [ M L 2 T − 2 ] [\mathrm{ML^{2}T^{-2}}] [ M L 2 T − 2 ] Show answer and solution
Answer: (d)
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Q20 JEE Main 2024 · 4 APR · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
Applying the principle of homogeneity of dimensions, determine which one is correct, where T T T is time period, G G G is gravitational constant, M M M is mass, r r r is radius of orbit.
(a) T 2 = 4 π 2 r G M 2 T^2 = \frac{4\pi^2 r}{GM^2} T 2 = G M 2 4 π 2 r (b) T 2 = 4 π 2 r 3 T^2 = 4\pi^2 r^3 T 2 = 4 π 2 r 3 (c) T 2 = 4 π 2 r 3 G M T^2 = \frac{4\pi^2 r^3}{GM} T 2 = GM 4 π 2 r 3 (d) T 2 = 4 π 2 r 2 G M T^2 = \frac{4\pi^2 r^2}{GM} T 2 = GM 4 π 2 r 2 Show answer and solution
Answer: (c)
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Q21 JEE Main 2024 · 4 APR · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
The equation of stationary wave is:
y = 2 a sin ( 2 π n t λ ) cos ( 2 π x λ ) y = 2\mathrm{a}\sin\left(\frac{2\pi\mathrm{nt}}{\lambda}\right)\cos\left(\frac{2\pi x}{\lambda}\right) y = 2 a sin ( λ 2 π nt ) cos ( λ 2 π x ) .
Which of the following is NOT correct:
(a) The dimensions of n t \mathrm{nt} nt is [L] (b) The dimensions of n n n is [ L T − 1 ] [\mathrm{LT^{-1}}] [ L T − 1 ] (c) The dimensions of x x x is [L] (d) The dimensions of n / λ n/\lambda n / λ is [T] Show answer and solution
Answer: (d)
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Q22 JEE Main 2024 · 31 JAN · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
Consider two physical quantities A A A and B B B related to each other as E = B − x 2 A t E = \frac{B - x^2}{At} E = A t B − x 2 where E , x E, x E , x and t t t have dimensions of energy, length and time respectively. The dimension of A B AB A B is
(a) L − 2 M 1 T 0 L^{-2}M^{1}T^{0} L − 2 M 1 T 0 (b) L 2 M − 1 T 1 L^{2}M^{-1}T^{1} L 2 M − 1 T 1 (c) L 0 M − 1 T 1 L^{0}M^{-1}T^{1} L 0 M − 1 T 1 (d) L − 2 M − 1 T 1 L^{-2}M^{-1}T^{1} L − 2 M − 1 T 1 Show answer and solution
Answer: (b)
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Q23 JEE Main 2024 · 31 JAN · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
A force is represented by F = a x 2 + b t 1 2 F = ax^2 + bt^{\frac{1}{2}} F = a x 2 + b t 2 1
where x = x = x = distance and t = t = t = time. The dimensions of b 2 / a b^2/a b 2 / a are:
(a) [ M L 2 T − 3 ] \left[\mathrm{ML^{2}T^{-3}}\right] [ M L 2 T − 3 ] (b) [ M L 3 T − 3 ] \left[\mathrm{ML^{3}T^{-3}}\right] [ M L 3 T − 3 ] (c) [ M L T − 2 ] [MLT^{-2}] [ M L T − 2 ] (d) [ M L − 1 T − 1 ] [ML^{-1}T^{-1}] [ M L − 1 T − 1 ] Show answer and solution
Answer: (b)
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Q24 JEE Main 2024 · 30 JAN · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
If mass is written as m = k c P G − 1 / 2 h 1 / 2 m = k\mathrm{c}^{\mathrm{P}}G^{-1/2}h^{1/2} m = k c P G − 1/2 h 1/2 then the value of P P P will be: (Constants have their usual meaning with k k k a dimensionless constant)
(a) 1/3 (b) − - − 1/3(c) 1/2 (d) 2 Show answer and solution
Answer: (c)
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Q25 JEE Main 2024 · 27 JAN · SHIFT 2 Medium ☆
Applications of Dimensional Analysis
The equation of state of a real gas is given by ( P + a V 2 ) ( V − b ) = R T \left(\mathrm{P} + \frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V} - \mathrm{b}) = \mathrm{RT} ( P + V 2 a ) ( V − b ) = RT , where P , V \mathrm{P, V} P , V and T \mathrm{T} T are pressure, volume and temperature respectively and R \mathrm{R} R is the universal gas constant. The dimensions of a b 2 \frac{\mathrm{a}}{\mathrm{b}^2} b 2 a is similar to that of:
Show answer and solution
Answer: (a)
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Q26 JEE Main 2023 · 13 APR · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
In the equation [ X + a Y 2 ] [ Y − b ] = R T , X \left[X + \frac{a}{Y^2}\right][Y - b] = \mathrm{R}T, X [ X + Y 2 a ] [ Y − b ] = R T , X is pressure, Y Y Y is volume, R \mathrm{R} R is universal gas constant and T T T is temperature. The physical quantity equivalent to the ratio a b \frac{a}{b} b a is:
(a) Impulse (b) Energy (c) Pressure gradient (d) Coefficient of viscosity Show answer and solution
Answer: (b)
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Q27 JEE Main 2023 · 11 APR · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
If force (F), velocity (V) and time (T) are considered as fundamental physical quantities, then dimensional formula of density will be:
(a) F V − 2 T 2 \mathrm{FV^{-2}T^{2}} F V − 2 T 2 (b) F V 4 T − 6 \mathrm{FV^{4}T^{-6}} F V 4 T − 6 (c) F 2 V − 2 T 6 \mathrm{F^{2}V^{-2}T^{6}} F 2 V − 2 T 6 (d) F V − 4 T − 2 \mathrm{FV^{-4}T^{-2}} F V − 4 T − 2 Show answer and solution
Answer: (d)
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Q28 JEE Main 2023 · 1 FEB · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
If the velocity of light c \mathrm{c} c , universal gravitational constant G \mathrm{G} G and Planck's constant h \mathrm{h} h are chosen as fundamental quantities. The dimensions of mass in the new system is:
(a) [ h 1 c 1 G − 1 ] \left[\mathrm{h^{1}c^{1}G^{-1}}\right] [ h 1 c 1 G − 1 ] (b) [ h − 1 / 2 c 1 / 2 G 1 / 2 ] \left[\mathrm{h^{-1/2}c^{1/2}G^{1/2}}\right] [ h − 1/2 c 1/2 G 1/2 ] (c) [ h 1 / 2 c 1 / 2 G − 1 / 2 ] \left[\mathrm{h^{1/2}c^{1/2}G^{-1/2}}\right] [ h 1/2 c 1/2 G − 1/2 ] (d) [ h 1 / 2 c − 1 / 2 G 1 ] \left[\mathrm{h^{1/2}c^{-1/2}G^{1}}\right] [ h 1/2 c − 1/2 G 1 ] Show answer and solution
Answer: (c)
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Q29 JEE Main 2023 · 1 FEB · SHIFT 1 Medium ☆
Applications of Dimensional Analysis
( P + a V 2 ) ( V − b ) = R T \left(P + \frac{a}{V^2}\right)(V - b) = RT ( P + V 2 a ) ( V − b ) = RT represents the equation of state of some gases. Where P P P is the pressure, V V V is the volume, T T T is the temperature and a , b , R a, b, R a , b , R are the constants. The physical quantity, which has dimensional formula as that of b 2 a \frac{b^2}{a} a b 2 , will be:
(a) Energy density (b) Bulk modulus (c) Modulus of rigidity (d) Compressibility Show answer and solution
Answer: (d)
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Q30 JEE Main 2023 · 29 JAN · SHIFT 2 Medium ☆
Applications of Dimensional Analysis
The equation of a circle is given by x 2 + y 2 = a 2 x^2 + y^2 = a^2 x 2 + y 2 = a 2 , where a is the radius. If the equation is modified to change the origin other than (0, 0), then find out the correct dimensions of A and B in a new equation: ( x − A t ) 2 + ( y − t B ) 2 = a 2 (x - At)^2 + \left(y - \frac{t}{B}\right)^2 = a^2 ( x − A t ) 2 + ( y − B t ) 2 = a 2 . The dimensions of t is given as [ T − 1 ] [\mathrm{T^{-1}}] [ T − 1 ] .
(a) A = [ L − 1 T − 1 ] , B = [ L T − 1 ] \mathrm{A} = [\mathrm{L^{-1}T^{-1}}], \mathrm{B} = [\mathrm{LT^{-1}}] A = [ L − 1 T − 1 ] , B = [ L T − 1 ] (b) A = [ L − 1 T − 1 ] , B = [ L T ] \mathrm{A} = [\mathrm{L^{-1}T^{-1}}], \mathrm{B} = [\mathrm{LT}] A = [ L − 1 T − 1 ] , B = [ LT ] (c) A = [ L T ] , B = [ L − 1 T − 1 ] \mathrm{A} = [\mathrm{LT}], \mathrm{B} = [\mathrm{L^{-1}T^{-1}}] A = [ LT ] , B = [ L − 1 T − 1 ] (d) A = [ L − 1 T ] , B = [ L T − 1 ] \mathrm{A} = [\mathrm{L^{-1}T}], \mathrm{B} = [\mathrm{LT^{-1}}] A = [ L − 1 T ] , B = [ L T − 1 ] Show answer and solution
Answer: (c)
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Q31 JEE Main 2023 · 29 JAN · SHIFT 1 Medium ☆
Applications of Dimensional Analysis
Match List I with List II:
Choose the correct answer from the options given below:
List I (Physical Quantity) List II (Dimensional Formula) A. Pressure gradient I. [ M 0 L 2 T − 2 ] \left[\mathrm{M^{0}L^{2}T^{-2}}\right] [ M 0 L 2 T − 2 ] B. Energy density II. [ M 1 L − 1 T − 2 ] \left[\mathrm{M^{1}L^{-1}T^{-2}}\right] [ M 1 L − 1 T − 2 ] C. Electric Field III. [ M 1 L − 2 T − 2 ] \left[\mathrm{M^{1}L^{-2}T^{-2}}\right] [ M 1 L − 2 T − 2 ] D. Latent heat IV. [ M 1 L 1 T − 3 A − 1 ] \left[\mathrm{M^{1}L^{1}T^{-3}A^{-1}}\right] [ M 1 L 1 T − 3 A − 1 ]
(a) A-III, B-II, C-IV, D-I (b) A-III, B-II, C-I, D-IV (c) A-II, B-III, C-IV, D-I (d) A-II, B-III, C-I, D-IV Show answer and solution
Answer: (a)
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Q32 JEE Main 2023 · 15 APR · SHIFT 1 Medium ☆
Applications of Dimensional Analysis
The speed of a wave produced in water is given by v = λ a g b ρ c v = \lambda^{a}g^{b}\rho^{c} v = λ a g b ρ c . Where λ , g \lambda, g λ , g and ρ \rho ρ are wavelength of wave, acceleration due to gravity and density of water respectively. The values of a , b a, b a , b and c c c respectively, are:
(a) 1 2 , 0 , 1 2 \frac{1}{2}, 0, \frac{1}{2} 2 1 , 0 , 2 1 (b) 1 , 1 , 0 1, 1, 0 1 , 1 , 0 (c) 1 , − 1 , 0 1,\ -1, 0 1 , − 1 , 0 (d) 1 2 , 1 2 , 0 \frac{1}{2}, \frac{1}{2}, 0 2 1 , 2 1 , 0 Show answer and solution
Answer: (d)
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Q33 JEE Main 2022 · 27 JUL · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
An expression of energy density is given by u = α β sin ( α x k t ) u = \frac{\alpha}{\beta}\sin\left(\frac{\alpha x}{kt}\right) u = β α sin ( k t αx ) , where α , β \alpha, \beta α , β are constants, x x x is displacement, k k k is Boltzmann constant and t is the temperature. The dimensions of β \beta β will be:
(a) [ M L 2 T − 2 θ − 1 ] \left[\mathrm{ML^{2}T^{-2}}\theta^{-1}\right] [ M L 2 T − 2 θ − 1 ] (b) [ M 0 L 2 T − 2 ] \left[\mathrm{M^{0}L^{2}T^{-2}}\right] [ M 0 L 2 T − 2 ] (c) [ M 0 L 0 T 0 ] \left[\mathrm{M^{0}L^{0}T^{0}}\right] [ M 0 L 0 T 0 ] (d) [ M 0 L 2 T 0 ] \left[\mathrm{M^{0}L^{2}T^{0}}\right] [ M 0 L 2 T 0 ] Show answer and solution
Answer: (d)
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Q34 JEE Main 2022 · 29 JUN · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
In van der Waal equation [ P + a V 2 ] \left[P + \frac{a}{V^2}\right] [ P + V 2 a ] [V − - − b] = RT; P is pressure, V is volume, R is universal gas constant and T is temperature. The ratio of constants a b \frac{a}{b} b a is dimensionally equal to:
(a) P V \frac{P}{V} V P (b) V P \frac{V}{P} P V (c) PV (d) PV3 Show answer and solution
Answer: (c)
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Q35 JEE Main 2022 · 30 JUL · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
If current (I), kinetic energy (K) and charge (Q) are taken as the fundamental quantities, The dimensional representation of power will be:
(a) [ K − 1 Q I − 1 ] [\mathrm{K^{-1}\,Q\,I^{-1}}] [ K − 1 Q I − 1 ] (b) [ K I Q − 1 ] [\mathrm{K\,I\,Q^{-1}}] [ K I Q − 1 ] (c) [ K Q I − 1 ] [\mathrm{K\,Q\,I^{-1}}] [ K Q I − 1 ] (d) [ K − 1 Q − 1 I ] [\mathrm{K^{-1}\,Q^{-1}\,I}] [ K − 1 Q − 1 I ] Show answer and solution
Answer: (b)
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Q36 JEE Main 2022 · 25 JUL · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
If momentum [P], area [ A ] [\mathrm{A}] [ A ] and time [ T ] [\mathrm{T}] [ T ] are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is:
(a) [ P A − 1 T 0 ] \left[\mathrm{P\,A^{-1}\,T^{0}}\right] [ P A − 1 T 0 ] (b) [ P A T − 1 ] \left[\mathrm{P\,A\,T^{-1}}\right] [ P A T − 1 ] (c) [ P A − 1 T ] [\mathrm{P\,A^{-1}\,T}] [ P A − 1 T ] (d) [ P A − 1 T − 1 ] \left[\mathrm{P\,A^{-1}\,T^{-1}}\right] [ P A − 1 T − 1 ] Show answer and solution
Answer: (a)
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Q37 JEE Main 2022 · 28 JUL · SHIFT 2 Medium ☆
Applications of Dimensional Analysis
Consider the efficiency of Carnot's engine is given by η = α β sin θ log e β x k T \eta = \frac{\alpha\beta}{\sin\theta}\log_e\frac{\beta x}{kT} η = s i n θ α β log e k T β x , where α \alpha α and β \beta β are constants. If T is temperature, k is Boltzmann constant, θ \theta θ is angular displacement and x has the dimensions of length. Then, choose the incorrect option:
(a) Dimensions of β \beta β is same as that of force. (b) Dimensions of α − 1 x \alpha^{-1}x α − 1 x is same as that of energy. (c) Dimensions of η − 1 sin θ \eta^{-1}\sin\theta η − 1 sin θ is same as that of α β \alpha\beta α β . (d) Dimensions of α \alpha α is same as that of β \beta β . Show answer and solution
Answer: (d)
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Q38 JEE Main 2022 · 28 JUN · SHIFT 2 Medium ☆
Applications of Dimensional Analysis
Velocity (v) and acceleration (a) in two systems of units 1 and 2 are related as v 2 = n m 2 v 1 v_2 = \frac{n}{m^2}v_1 v 2 = m 2 n v 1 and a 2 = a 1 m n a_2 = \frac{a_1}{mn} a 2 = mn a 1 respectively. Here m and n are constants. The relations for distance and time in two systems respectively are:
(a) n 3 m 3 L 1 = L 2 \frac{n^3}{m^3}L_1 = L_2 m 3 n 3 L 1 = L 2 and n 2 m T 1 = T 2 \frac{n^2}{m}T_1 = T_2 m n 2 T 1 = T 2 (b) L 1 = n 4 m 2 L 2 L_1 = \frac{n^4}{m^2}L_2 L 1 = m 2 n 4 L 2 and T 1 = n 2 m T 2 T_1 = \frac{n^2}{m}T_2 T 1 = m n 2 T 2 (c) L 1 = n 2 m L 2 L_1 = \frac{n^2}{m}L_2 L 1 = m n 2 L 2 and T 1 = n 4 m 2 T 2 T_1 = \frac{n^4}{m^2}T_2 T 1 = m 2 n 4 T 2 (d) n 2 m L 1 = L 2 \frac{n^2}{m}L_1 = L_2 m n 2 L 1 = L 2 and n 4 m 2 T 1 = T 2 \frac{n^4}{m^2}T_1 = T_2 m 2 n 4 T 1 = T 2 Show answer and solution
Answer: (a)
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Q39 JEE Main 2022 · 26 JUN · SHIFT 1 Medium ☆
Applications of Dimensional Analysis
An expression for a dimensionless quantity P is given by P = α β log e ( k t β x ) P = \frac{\alpha}{\beta}\log_e\left(\frac{kt}{\beta x}\right) P = β α log e ( β x k t ) ; where α \alpha α and β \beta β are constants, x is distance; k is Boltzmann constant and t is the temperature. Then the dimensions of α \alpha α will be:
(a) [M0 L− 1 {}^{-1} − 1 T0] (b) [M L0 T− 2 {}^{-2} − 2 ] (c) [M L T− 2 {}^{-2} − 2 ] (d) [M L2 T− 2 {}^{-2} − 2 ] Show answer and solution
Answer: (c)
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Q40 JEE Main 2021 · 25 JUL · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
The force is given in terms of time t and displacement x by the equation
F = A cos Bx + C sin Dt
The dimensional formula of A D B \frac{AD}{B} B A D is:
(a) [ M 0 L T − 1 ] [M^0LT^{-1}] [ M 0 L T − 1 ] (b) [ M L 2 T − 3 ] [ML^2T^{-3}] [ M L 2 T − 3 ] (c) [ M 1 L 1 T − 2 ] [M^1L^1T^{-2}] [ M 1 L 1 T − 2 ] (d) [ M 2 L 2 T − 3 ] [M^2L^2T^{-3}] [ M 2 L 2 T − 3 ] Show answer and solution
Answer: (b)
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Q41 NEET 2021 Easy ☆
Applications of Dimensional Analysis
If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.
(a) [F] [A− 1 {}^{-1} − 1 ] [T] (b) [F] [A] [T] (c) [F] [A] [T2] (d) [F] [A] [T− 1 {}^{-1} − 1 ] Show answer and solution
Answer: (c)
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Q42 JEE Main 2021 · 31 AUG · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
If velocity [V], time [T] and force [F] are chosen as the base quantities, the dimensions of the mass will be:
(a) [FT− 1 {}^{-1} − 1 V− 1 {}^{-1} − 1 ] (b) [FTV− 1 {}^{-1} − 1 ] (c) [FT2 V] (d) [FVT− 1 {}^{-1} − 1 ] Show answer and solution
Answer: (b)
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Q43 JEE Main 2021 · 27 AUG · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
If force (F), length (L) and time (T) are taken as the fundamental quantities. Then what will be the dimension of density:
(a) [FL− 4 {}^{-4} − 4 T2] (b) [FL− 3 {}^{-3} − 3 T2] (c) [FL− - − 5T2] (d) [FL− 3 {}^{-3} − 3 T3] Show answer and solution
Answer: (a)
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Q44 JEE Main 2021 · 20 JUL · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
If time (t), velocity (v), and angular momentum (l) are taken as the fundamental units. Then the dimension of mass (m) in terms of t, v and l is:
(a) [ t − 1 v 1 l − 2 ] [t^{-1}v^{1}l^{-2}] [ t − 1 v 1 l − 2 ] (b) [ t 1 v 2 l − 1 ] [t^{1}v^{2}l^{-1}] [ t 1 v 2 l − 1 ] (c) [ t − 2 v − 1 l 1 ] [t^{-2}v^{-1}l^{1}] [ t − 2 v − 1 l 1 ] (d) [ t − 1 v − 2 l 1 ] [t^{-1}v^{-2}l^{1}] [ t − 1 v − 2 l 1 ] Show answer and solution
Answer: (d)
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Q45 JEE Main 2021 · 26 AUG · SHIFT 1 Medium ☆
Applications of Dimensional Analysis
If E, L, M and G denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of P in the formula P = EL2M− 5 G − -5\mathrm{G}- − 5 G − 2 are:
(a) [M0 L1 T0] (b) [M− 1 {}^{-1} − 1 L− 1 {}^{-1} − 1 T2] (c) [M1 L1 T− 2 {}^{-2} − 2 ] (d) [M0 L0 T0] Show answer and solution
Answer: (d)
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Q46 JEE Main 2021 · 20 JUL · SHIFT 1 Medium ☆
Applications of Dimensional Analysis
The entropy of any system is given by
S = α 2 β ln [ μ k R J β 2 + 3 ] S = \alpha^2\beta\ln\left[\frac{\mu kR}{J\beta^2} + 3\right] S = α 2 β ln [ J β 2 μ k R + 3 ] where α \alpha α and β \beta β are the constants. μ \mu μ , J, k and R are no. of moles, mechanical equivalent of heat, Boltzmann constant and gas constant respectively.
[Take S = d Q T S = \frac{dQ}{T} S = T d Q ]
Choose the incorrect option from the following:
(a) α \alpha α and J have the same dimensions.(b) S and α \alpha α have different dimensions (c) S, β \beta β , k and μ \mu μ R have the same dimensions (d) α \alpha α and k have the same dimensionsShow answer and solution
Answer: (d)
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Q47 JEE Main 2021 · 26 FEB · SHIFT 1 Medium ☆
Applications of Dimensional Analysis
In a typical combustion engine the workdone by a gas molecule is given by W = α 2 β e − β x 2 k T W = \alpha^2\beta e^{\frac{-\beta x^2}{kT}} W = α 2 β e k T − β x 2 , where x is the displacement, k is the Boltzmann constant and T is the temperature. If α \alpha α and β \beta β are constants, dimensions of α \alpha α will be:
(a) [ M 0 L T 0 ] [M^0LT^0] [ M 0 L T 0 ] (b) [ M L T − 1 ] [MLT^{-1}] [ M L T − 1 ] (c) [ M L T − 2 ] [MLT^{-2}] [ M L T − 2 ] (d) [ M 2 L T − 2 ] [M^2LT^{-2}] [ M 2 L T − 2 ] Show answer and solution
Answer: (a)
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Q48 JEE Main 2021 · 24 FEB · SHIFT 1 Medium ☆
Applications of Dimensional Analysis
The work done by a gas molecule in an isolated system is
given by, W = α β 2 e − x 2 α k T W = \alpha\beta^2 e^{-\frac{x^2}{\alpha kT}} W = α β 2 e − α k T x 2 , where x is the displacement, k is the Boltzmann constant and T is the temperature. α \alpha α and β \beta β are constants. Then the dimensions of β \beta β will be:
(a) [ M 0 L T 0 ] [M^0LT^0] [ M 0 L T 0 ] (b) [ M L 2 T − 2 ] [ML^2T^{-2}] [ M L 2 T − 2 ] (c) [ M L T − 2 ] [MLT^{-2}] [ M L T − 2 ] (d) [ M 2 L T 2 ] [M^2LT^2] [ M 2 L T 2 ] Show answer and solution
Answer: (c)
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Q49 JEE Main 2021 · 31 AUG · SHIFT 1 Difficult ☆
Applications of Dimensional Analysis
Which of the following equations is dimensionally incorrect?
Where t = time, h = height, s = surface tension, θ \theta θ = angle, ρ \rho ρ = density, a, r = radius, g = acceleration due to gravity, v = volume, p = pressure, W = work done, Γ \Gamma Γ = torque, ϵ \epsilon ϵ = permittivity, E = electric field, J = current density, L = length.
(a) v = π p a 4 8 η L v = \frac{\pi p a^4}{8\eta L} v = 8 η L π p a 4 (b) h = 2 s cos θ ρ r g h = \frac{2s\cos\theta}{\rho r g} h = ρ r g 2 s c o s θ (c) J = ϵ ∂ E ∂ t J = \epsilon \frac{\partial E}{\partial t} J = ϵ ∂ t ∂ E (d) W = Γ θ W = \Gamma\theta W = Γ θ Show answer and solution
Answer: (a)
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Q50 MHT CET 2020 · 16 OCT · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
Let force F = A sin ( C t ) + B cos ( D x ) F = A\sin(Ct) + B\cos(Dx) F = A sin ( Ct ) + B cos ( D x ) , where x x x and t t t are displacement and time, respectively. The dimensions of C D \frac{C}{D} D C are same as dimensions of
(a) angular velocity (b) velocity (c) angular momentum (d) velocity gradient Show answer and solution
Answer: (b)
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Q51 JEE Main 2020 · 2 SEP · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
If momentum (P), area (A) and time (T) are taken to be the fundamental quantities then the dimensional formula for energy is
(a) [P2AT–2] (b) [ P 1 2 A T − 1 ] \left[P^{\frac{1}{2}}AT^{-1}\right] [ P 2 1 A T − 1 ] (c) [ P A 1 2 T − 1 ] \left[PA^{\frac{1}{2}}T^{-1}\right] [ P A 2 1 T − 1 ] (d) [PA–1T–2] Show answer and solution
Answer: (c)
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Q52 JEE Main 2020 · 8 JAN · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
The dimension of stopping potential V0 in photoelectric effect in units of Planck's constant 'h', speed of light 'c' and Gravitational constant 'G' and ampere A is:
(a) h1/3 G2/3 c1/3 A–1 (b) h0 c5 G-1 A-1 (c) h2/3 c5/3 G1/3 A–1 (d) h2 G3/2 c1/3 A–1 Show answer and solution
Answer: (b)
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Q53 JEE Main 2020 · 4 SEP · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
A quantity x is given by ( I F v 2 W L 4 ) \left(\frac{IFv^2}{WL^4}\right) ( W L 4 I F v 2 ) in terms of moment of inertia I, force F, velocity v, work W and Length L. The dimensional formula for x is same as that of:
(a) Coefficient of viscosity (b) Force constant (c) Energy density (d) Planck's constant Show answer and solution
Answer: (c)
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Q54 JEE Main 2020 · 2 SEP · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
If speed V, area A and force F are chosen as fundamental units, then the dimension of Young’s modulus will be
(a) FA–1V0 (b) FA2V–1 (c) FA2V–2 (d) FA2V–3 Show answer and solution
Answer: (a)
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Q55 JEE Main 2020 · 9 JAN · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
A quantity f is given by f = h c 5 G f = \sqrt{\frac{hc^5}{G}} f = G h c 5 where c is speed of light, G universal gravitational constant and h is the Planck's constant. Dimension of f is that of:
(a) Energy (b) Momentum (c) Area (d) Volume Show answer and solution
Answer: (a)
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Q56 MHT CET 2019 · 2 MAY · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
The force ' F F F ' acting on a body of density ' d d d ' are related by the relation F = y d F = \frac{y}{\sqrt{d}} F = d y . The dimensions of ' y y y ' are
(a) [ L − 1 2 M 3 2 T − 2 ] \left[\mathrm{L^{-\frac{1}{2}}M^{\frac{3}{2}}T^{-2}}\right] [ L − 2 1 M 2 3 T − 2 ] (b) [ L − 1 M 1 2 T − 2 ] \left[L^{-1}M^{\frac{1}{2}}T^{-2}\right] [ L − 1 M 2 1 T − 2 ] (c) [ L − 1 M 3 2 T − 2 ] \left[L^{-1}M^{\frac{3}{2}}T^{-2}\right] [ L − 1 M 2 3 T − 2 ] (d) [ L − 1 2 M 1 2 T − 2 ] \left[\mathrm{L^{-\frac{1}{2}}M^{\frac{1}{2}}T^{-2}}\right] [ L − 2 1 M 2 1 T − 2 ] Show answer and solution
Answer: (a)
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Q57 JEE Main 2019 · 10 APR · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
In the formula X = 5YZ2, X and Z have dimensions of capacitance and magnetic field, respectively. What are the dimensions of Y in SI units?
(a) [M–3L–2T8A4] (b) [M–2L–2T6A3] (c) [M–1L–2T4A2] (d) [M–2L0 T–4A–2] Show answer and solution
Answer: (a)
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Q58 JEE Main 2019 · 11 JAN · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young's modulus will be:
(a) V− 2 {}^{-2} − 2 A2F2 (b) V− 4 A − -4\mathrm{A}- − 4 A − 2F (c) V− 4 {}^{-4} − 4 A2F (d) V− 2 {}^{-2} − 2 A2F− 2 {}^{-2} − 2 Show answer and solution
Answer: (c)
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Q59 JEE Main 2019 · 9 JAN · SHIFT 2 Easy ☆
Applications of Dimensional Analysis
Expression for time in terms of G (universal gravitational constant), h (Planck constant) and c (speed of light) is proportional to:
(a) h c 5 G \sqrt{\frac{hc^5}{G}} G h c 5 (b) c 3 G h \sqrt{\frac{c^3}{Gh}} G h c 3 (c) G h c 5 \sqrt{\frac{Gh}{c^5}} c 5 G h (d) G h c 3 \sqrt{\frac{Gh}{c^3}} c 3 G h Show answer and solution
Answer: (c)
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Q60 JEE Main 2019 · 11 JAN · SHIFT 1 Medium ☆
Applications of Dimensional Analysis
The force of interaction between two atoms is given by F = α β \alpha\beta α β exp ( − x 2 α k T ) \left(-\frac{x^2}{\alpha kT}\right) ( − α k T x 2 ) ; where x is the distance, k is the Boltzmann constant and T is temperature and α \alpha α and β \beta β are two constants. The dimension of β \beta β is:
(a) M2L2T− 2 {}^{-2} − 2 (b) M2LT− 4 {}^{-4} − 4 (c) MLT− 4 {}^{-4} − 4 (d) M0L2T− 4 {}^{-4} − 4 Show answer and solution
Answer: (b)
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Q61 JEE Main 2019 · 8 APR · SHIFT 2 Medium ☆
Applications of Dimensional Analysis
If surface tension (S), Moment of inertia (I) and Planck's constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be :-
(a) S1/2I1/2h0 (b) S3/2I1/2h0 (c) S1/2I1/2h-1 (d) S1/2I3/2h-1 Show answer and solution
Answer: (a)
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Q62 JEE Main 2019 · 10 JAN · SHIFT 1 Medium ☆
Applications of Dimensional Analysis
The density of a material in SI units is 128 kg m–3. In certain units in which the unit of length is 25 cm and the unit of mass is 50 g, the numerical value of density of the material is -
Show answer and solution
Answer: (a)
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Q63 JEE Main 2018 · 15 APR · SHIFT 2 Medium ☆
Applications of Dimensional Analysis
The characteristic distance at which quantum gravitational effects are significant, the Planck length, can be determined from a suitable combination of the fundamental physical constants G, h and c. Which of the following correctly gives the Planck length?
(a) G ℏ \hbar ℏ 2 c3 (b) G2 ℏ \hbar ℏ c (c) G 1 2 ℏ 2 c G^{\frac{1}{2}}\hbar^2 c G 2 1 ℏ 2 c (d) ( G ℏ c 3 ) 1 2 \left(\frac{G\hbar}{c^3}\right)^{\frac{1}{2}} ( c 3 G ℏ ) 2 1 Show answer and solution
Answer: (d)
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Q64 JEE Main 2017 · 8 APR · SHIFT 1 Easy ☆
Applications of Dimensional Analysis
Time (T), velocity (C) and angular momentum (h) are chosen as fundamentalquantities instead of mass, length and time. In terms of these, the dimensions of mass would be:
(a) [M] = [T− 1 {}^{-1} − 1 C− 2 {}^{-2} − 2 h] (b) [M] = [T− 1 {}^{-1} − 1 C2 h] (c) [M] = [T− 1 {}^{-1} − 1 C− 2 {}^{-2} − 2 h− 1 {}^{-1} − 1 ] (d) [M] = [T C− 2 {}^{-2} − 2 h] Show answer and solution
Answer: (a)
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Q65 NEET 2017 Medium ☆
Applications of Dimensional Analysis
A physical quantity of the dimensions of length that can be formed out of c, G and e 2 4 π ε 0 \frac{e^2}{4\pi\varepsilon_0} 4 π ε 0 e 2 is [c is velocity of light, G is the universal constant of gravitation and e is charge]
(a) c 2 [ G − e 2 4 π ε 0 ] 1 / 2 c^2\left[G - \frac{e^2}{4\pi\varepsilon_0}\right]^{1/2} c 2 [ G − 4 π ε 0 e 2 ] 1/2 (b) 1 c 2 [ e 2 G 4 π ε 0 ] 1 / 2 \frac{1}{c^2}\left[\frac{e^2}{G\,4\pi\varepsilon_0}\right]^{1/2} c 2 1 [ G 4 π ε 0 e 2 ] 1/2 (c) 1 c G e 2 4 π ε 0 \frac{1}{c}G\frac{e^2}{4\pi\varepsilon_0} c 1 G 4 π ε 0 e 2 (d) 1 c 2 [ G e 2 4 π ε 0 ] 1 / 2 \frac{1}{c^2}\left[G\frac{e^2}{4\pi\varepsilon_0}\right]^{1/2} c 2 1 [ G 4 π ε 0 e 2 ] 1/2 Show answer and solution
Answer: (d)
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Q66 JEE Main 2016 · 10 APR · SHIFT 1 (One or more correct) Medium ☆
Applications of Dimensional Analysis
A, B, C and D are four different physical quantities having different dimensions. None of them is dimensionless. But we know that the equation AD = C ln (BD) holds true. Then which of the combination is not a meaningful quantity?
(a) A2 − - − B2C2 (b) ( A − C ) D \frac{(A-C)}{D} D ( A − C ) (c) A B − C \frac{A}{B} - C B A − C (d) C B D − A D 2 C \frac{C}{BD} - \frac{AD^2}{C} B D C − C A D 2 Show answer and solution
Answer: (b)
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Q67 NEET 2016 · PHASE 2 Medium ☆
Applications of Dimensional Analysis
Planck's constant (h), speed of light in vacuum (c) and Newton's gravitional constant (G) are three fundamental constants. Which of the following combinations of these has the dimension of length?
(a) h G c 3 / 2 \frac{\sqrt{hG}}{c^{3/2}} c 3/2 h G (b) h G c 5 / 2 \frac{\sqrt{hG}}{c^{5/2}} c 5/2 h G (c) h c G \sqrt{\frac{hc}{G}} G h c (d) G c h 3 / 2 \sqrt{\frac{Gc}{h^{3/2}}} h 3/2 G c Show answer and solution
Answer: (a)
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Q68 NEET 2015 · AIPMT · CANCELLED PAPER Easy ☆
Applications of Dimensional Analysis
If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be
(a) [ E V − 2 T − 2 ] [EV^{-2}T^{-2}] [ E V − 2 T − 2 ] (b) [ E − 2 V − 1 T − 3 ] [E^{-2}V^{-1}T^{-3}] [ E − 2 V − 1 T − 3 ] (c) [ E V − 2 T − 1 ] [EV^{-2}T^{-1}] [ E V − 2 T − 1 ] (d) [ E V − 1 T − 2 ] [EV^{-1}T^{-2}] [ E V − 1 T − 2 ] Show answer and solution
Answer: (a)
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Q69 JEE Main 2015 · 11 APR · SHIFT 1 Medium ☆
Applications of Dimensional Analysis
If electronic charge e, electron mass m, speed of light in vacuum c and Planck's constant h are taken as fundamental quantities, the permeability of vacuum μ \mu μ 0 can be expressed in units of:
(a) ( h c m e 2 ) \left(\frac{hc}{me^2}\right) ( m e 2 h c ) (b) ( h m e 2 ) \left(\frac{h}{me^2}\right) ( m e 2 h ) (c) ( h c e 2 ) \left(\frac{h}{ce^2}\right) ( c e 2 h ) (d) ( m c 2 h e 2 ) \left(\frac{mc^2}{he^2}\right) ( h e 2 m c 2 ) Show answer and solution
Answer: (c)
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Q70 JEE Main 2015 · 10 APR · SHIFT 1 Medium ☆
Applications of Dimensional Analysis
If the capacitance of a nanocapacitor is measured in terms of a unit 'u' made by combining the electronic charge 'e', Bohr radius 'a0 ', Planck's constant 'h' and speed of light 'c' then:
(a) u = e 2 c h a 0 \frac{e^2c}{ha_0} h a 0 e 2 c (b) u = e 2 h c a 0 \frac{e^2h}{ca_0} c a 0 e 2 h (c) u = e 2 a 0 h c \frac{e^2a_0}{hc} h c e 2 a 0 (d) u = h c e 2 a 0 \frac{hc}{e^2a_0} e 2 a 0 h c Show answer and solution
Answer: (c)
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Q71 NEET 2015 · AIPMT Medium ☆
Applications of Dimensional Analysis
If dimensions of critical velocity v v v c of a liquid flowing through a tube are expressed as [ η x ρ y r z ] [\eta^{x}\rho^{y}r^{z}] [ η x ρ y r z ] where η , ρ \eta, \rho η , ρ and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by
(a) − - − 1, − - − 1, − - − 1(b) 1, 1, 1 (c) 1, − - − 1, − - − 1 (d) − - − 1, − - − 1, 1Show answer and solution
Answer: (c)
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Q72 JEE Main 2015 · 11 APR · SHIFT 1 Difficult ☆
Applications of Dimensional Analysis
A beaker contains a fluid of density ρ \rho ρ kg / m3, specific heat S J/kgoC and viscosity η \eta η . The beaker is filled up to height h. To estimate the rate of heat transfer per unit area ( Q ˙ / A \dot{Q}/A Q ˙ / A ) by convection when beaker is put on a hot plate, a student proposes that it should depend on η ⋅ ( S Δ θ h ) \eta \cdot \left(\frac{S\Delta\theta}{h}\right) η ⋅ ( h S Δ θ ) and ( 1 ρ g ) \left(\frac{1}{\rho g}\right) ( ρ g 1 ) when Δ θ \Delta\theta Δ θ (inoC) is the difference in the temperature between the bottom and top of the fluid. In that situation the correct option for (Q ˙ / A \dot{Q}/A Q ˙ / A ) is:
(a) η S Δ θ h \eta\frac{S\Delta\theta}{h} η h S Δ θ (b) η ( S Δ θ h ) ( 1 ρ g ) \eta\left(\frac{S\Delta\theta}{h}\right)\left(\frac{1}{\rho g}\right) η ( h S Δ θ ) ( ρ g 1 ) (c) S Δ θ η h \frac{S\Delta\theta}{\eta h} η h S Δ θ (d) ( S Δ θ η h ) ( 1 ρ g ) \left(\frac{S\Delta\theta}{\eta h}\right)\left(\frac{1}{\rho g}\right) ( η h S Δ θ ) ( ρ g 1 ) Show answer and solution
Answer: (a)
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Q73 NEET 2014 · AIPMT Easy ☆
Applications of Dimensional Analysis
If force (F), velocity (V) and time (T) are taken as fundamental units, then dimensions of mass are
(a) [ F V T − 1 ] [FVT^{-1}] [ F V T − 1 ] (b) [ F V T − 2 ] [FVT^{-2}] [ F V T − 2 ] (c) [ F V − 1 T − 1 ] [FV^{-1}T^{-1}] [ F V − 1 T − 1 ] (d) [ F V − 1 T ] [FV^{-1}T] [ F V − 1 T ] Show answer and solution
Answer: (d)
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Q74 JEE Main 2013 · 9 APR · SHIFT 1 Medium ☆
Applications of Dimensional Analysis
From the following, the quantity (constructed from the basic constants of nature), that has the dimensions, as well as correct order of magnitude, vis-a-vis typical atomic size, is:
(a) e 2 4 π ϵ 0 m e c 2 \frac{e^2}{4\pi\epsilon_0 m_e c^2} 4 π ϵ 0 m e c 2 e 2 (b) 4 π ϵ 0 e 2 m e 2 \frac{4\pi\epsilon_0 e^2}{m_e^2} m e 2 4 π ϵ 0 e 2 (c) m e 4 4 π ϵ 0 ℏ 2 \frac{me^4}{4\pi\epsilon_0\hbar^2} 4 π ϵ 0 ℏ 2 m e 4 (d) 4 π ϵ 0 m e c 2 e 4 \frac{4\pi\epsilon_0 m_e c^2}{e^4} e 4 4 π ϵ 0 m e c 2 Show answer and solution
Answer: (a)
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Q75 JEE Main 2013 · 25 APR · SHIFT 1 Medium ☆
Applications of Dimensional Analysis
A metal sample carrying a current along X-axis with density J x J_x J x is subjected to a magnetic field B z B_z B z (along z-axis). The electric field E y E_y E y developed along Y-axis is directly proportional to J x J_x J x as well as B z B_z B z . The constant of proportionality has SI unit.
(a) m 2 A \frac{m^2}{A} A m 2 (b) m 3 A s \frac{m^3}{As} A s m 3 (c) m 2 A s \frac{m^2}{As} A s m 2 (d) A s m 3 \frac{As}{m^3} m 3 A s Show answer and solution
Answer: (b)
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Q76 JEE Main 2013 · 23 APR · SHIFT 1 Medium ☆
Applications of Dimensional Analysis
If the time period t t t of the oscillation of a drop of liquid of density d d d , radius r r r , vibrating under surface tension s s s is given by the formula t = r 2 b s c d a / 2 t = \sqrt{r^{2b}s^{c}d^{a/2}} t = r 2 b s c d a /2 . It is observed that the time period is directly proportional to d s \sqrt{\frac{d}{s}} s d . The value of b b b should therefore be:
(a) 3 4 \frac{3}{4} 4 3 (b) 3 \sqrt{3} 3 (c) 3 2 \frac{3}{2} 2 3 (d) 2 3 \frac{2}{3} 3 2 Show answer and solution
Answer: (c)
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Q77 NEET 2011 · AIPMT · MAINS Medium ☆
Applications of Dimensional Analysis
The density of a material in CGS system of units is 4 g cm− 3 {}^{-3} − 3 . In a system of units in which unit of length is 10 cm and unit of mass is 100 g, the value of density of material will be
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Answer: (c)
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Q78 NEET 2010 · AIPMT · PRELIMS Easy ☆
Applications of Dimensional Analysis
The dimension of 1 2 ε 0 E 2 \frac{1}{2}\varepsilon_0 E^2 2 1 ε 0 E 2 , where ε 0 \varepsilon_0 ε 0 is permittivity of free space and E is electric field, is
(a) ML2T− 2 {}^{-2} − 2 (b) ML− 1 T − {}^{-1}\mathrm{T}^{-} − 1 T − 2 (c) ML2T− 1 {}^{-1} − 1 (d) MLT− 1 {}^{-1} − 1 Show answer and solution
Answer: (b)
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Q79 NEET 2006 · AIPMT Easy ☆
Applications of Dimensional Analysis
The velocity v of a particle at time t is given by v = at + b t + c \frac{b}{t + c} t + c b , where a, b and c are constants. The dimensions of a, b and c are
(a) [L], [LT] and [LT− 2 {}^{-2} − 2 ] (b) [LT− 2 {}^{-2} − 2 ], [L] and [T] (c) [L2], [T] and [LT− 2 {}^{-2} − 2 ] (d) [LT− 2 {}^{-2} − 2 ], and [LT] and [L] Show answer and solution
Answer: (b)
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Q80 NEET 1996 · AIPMT Easy ☆
Applications of Dimensional Analysis
An equation is given here ( P + a V 2 ) = b θ V \left(P + \frac{a}{V^2}\right) = b\frac{\theta}{V} ( P + V 2 a ) = b V θ where P P P = Pressure, V V V = Volume and θ \theta θ = Absolute temperature. If a a a and b b b are constants, the dimensions of a a a will be
(a) [ M L − 5 T − 1 ] [\mathrm{ML^{-5}T^{-1}}] [ M L − 5 T − 1 ] (b) [ M L 5 T 1 ] [\mathrm{ML^{5}T^{1}}] [ M L 5 T 1 ] (c) [ M L 5 T − 2 ] [\mathrm{ML^{5}T^{-2}}] [ M L 5 T − 2 ] (d) [ M − 1 L 5 T 2 ] [\mathrm{M^{-1}L^{5}T^{2}}] [ M − 1 L 5 T 2 ] Show answer and solution
Answer: (c)
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Q81 NEET 1993 · AIPMT Easy ☆
Applications of Dimensional Analysis
The time dependence of a physical quantity p p p is given by p = p 0 exp ( − a t 2 ) p = p_0 \exp(-at^2) p = p 0 exp ( − a t 2 ) , where a a a is a constant and t t t is the time. The constant a a a
(a) is dimensionless (b) has dimensions [ T − 2 ] [\mathrm{T^{-2}}] [ T − 2 ] (c) has dimensions [ T 2 ] [\mathrm{T^{2}}] [ T 2 ] (d) has dimensions of p p p Show answer and solution
Answer: (b)
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Q82 NEET 1993 · AIPMT Medium ☆
Applications of Dimensional Analysis
Turpentine oil is flowing through a tube of length l l l and radius r r r . The pressure difference between the two ends of the tube is P P P . The viscosity of oil is given by η = P ( r 2 − x 2 ) 4 v l \eta = \frac{P(r^2 - x^2)}{4vl} η = 4 v l P ( r 2 − x 2 ) where v v v is the velocity of oil at a distance x x x from the axis of the tube. The dimensions of η \eta η are
(a) [ M 0 L 0 T 0 ] [\mathrm{M^{0}L^{0}T^{0}}] [ M 0 L 0 T 0 ] (b) [ M L T − 1 ] [\mathrm{MLT^{-1}}] [ ML T − 1 ] (c) [ M L 2 T − 2 ] [\mathrm{ML^{2}T^{-2}}] [ M L 2 T − 2 ] (d) [ M L − 1 T − 1 ] [\mathrm{ML^{-1}T^{-1}}] [ M L − 1 T − 1 ] Show answer and solution
Answer: (d)
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Q83 NEET 1990 · AIPMT Easy ☆
Applications of Dimensional Analysis
The frequency of vibration f f f of a mass m m m suspended from a spring of spring constant k k k is given by a relation f = a m x k y f = am^xk^y f = a m x k y , where a a a is a dimensionless constant. The values of x x x and y y y are
(a) x = 1 2 , y = 1 2 x = \frac{1}{2}, y = \frac{1}{2} x = 2 1 , y = 2 1 (b) x = − 1 2 , y = − 1 2 x = -\frac{1}{2}, y = -\frac{1}{2} x = − 2 1 , y = − 2 1 (c) x = 1 2 , y = − 1 2 x = \frac{1}{2}, y = -\frac{1}{2} x = 2 1 , y = − 2 1 (d) x = − 1 2 , y = 1 2 x = -\frac{1}{2}, y = \frac{1}{2} x = − 2 1 , y = 2 1 Show answer and solution
Answer: (d)
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Q84 NEET 1990 · AIPMT Medium ☆
Applications of Dimensional Analysis
According to Newton, the viscous force acting between liquid layers of area A A A and velocity gradient Δ v / Δ Z \Delta v/\Delta Z Δ v /Δ Z is given by F = − η A Δ v Δ Z F = -\eta A\frac{\Delta v}{\Delta Z} F = − η A Δ Z Δ v , where η \eta η is constant called coefficient of viscosity. The dimensional formula of η \eta η is
(a) [ M L − 2 T − 2 ] [\mathrm{ML^{-2}T^{-2}}] [ M L − 2 T − 2 ] (b) [ M 0 L 0 T 0 ] [\mathrm{M^{0}L^{0}T^{0}}] [ M 0 L 0 T 0 ] (c) [ M L 2 T − 2 ] [\mathrm{ML^{2}T^{-2}}] [ M L 2 T − 2 ] (d) [ M L − 1 T − 1 ] [\mathrm{ML^{-1}T^{-1}}] [ M L − 1 T − 1 ] Show answer and solution
Answer: (d)
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