 JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2004

Which one of the following represents the correct dimensions of the coefficient of viscosity?
A
ML-1T-1
B
MLT-1
C
ML-1T-2
D
ML-2T-2

Explanation

From stokes law, Viscous force F = $6\pi \eta rv$

$\Rightarrow \eta = {F \over {6\pi rv}}$

$\therefore$ $[\eta] = {[{ML{T^{ - 2}}]} \over {[L][L{T^{ - 1}}]}}$

$\Rightarrow [\eta] = [M{L^{ - 1}}{T^{ - 1}}]$
2

AIEEE 2003

The physical quantities not having same dimensions are
A
torque and work
B
momentum and Planck's constant
C
stress and Young's modulus
D
speed and ${\left( {{\mu _0}{\varepsilon _0}} \right)^{ - 1/2}}$

Explanation

Momentum = mv = [${M{L}{T^{ - 1}}}$]

Planck's constant, h = ${E \over v}$ = ${[{M{L^2}{T^{ - 2}]}} \over {[{T^{ - 1}}]}}$ = $[{M{L^2}{T^{ - 1}}}]$

So Momentum and Planck's constant do not have same dimensions.
3

AIEEE 2003

Dimensions of ${1 \over {{\mu _0}{\varepsilon _0}}}$, where symbols have their usual meaning, are
A
[ L-1T ]
B
[ L-2T2 ]
C
[ L2T-2 ]
D
[ LT-1 ]

Explanation

The velocity of light in vacuum is

c = ${1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$;

$\therefore[{1 \over {{\mu _0}{\varepsilon _0}}}]$ = [c2] = [L2T-2]

$\therefore$ Dimension of ${1 \over {{\mu _0}{\varepsilon _0}}}$ = [L2T-2]
4

AIEEE 2002

Identify the pair whose dimensions are equal
A
torque and work
B
stress and energy
C
force and stress
D
force and work

Explanation

Work (W) = $\overrightarrow F .\overrightarrow s$ = Fs cos$\theta$

= [MLT-2][L] = [ML2T-2];

Torque ($\overrightarrow \tau) = \overrightarrow r \times \overrightarrow F$ $\Rightarrow \tau = rF\sin \theta$

= [L] [MLT-2] = [ML2T-2]

So dimension of torque and work are same.