1

IIT-JEE 2010

MCQ (More than One Correct Answer)
A student uses a simple pendulum of exactly 1m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true?
A
Error ΔT in measuring T, the time period, is 0.05 seconds
B
Error ΔT in measuring T, the time period, is 1 second
C
Percentage error in the determination of g is 5%
D
Percentage error in the determination of g is 2.5%
2

IIT-JEE 1998

MCQ (More than One Correct Answer)
Let [$${\mathrm\varepsilon}_\mathrm o$$] denote the dimentional formula of the permittivity of the vacuum, and [$$\mu_o$$] that of the permeability of the vacuum.If M = mass, L = length, T = time and I = electric current,
A
$$\left[{\mathrm\varepsilon}_\mathrm o\right]\;=\;\mathrm M^{-1}\mathrm L^{-3}\mathrm T^2\;\mathrm I$$
B
$$\left[{\mathrm\varepsilon}_\mathrm o\right]\;=\;\mathrm M^{-1}\mathrm L^{-3}\mathrm T^4\mathrm I^2$$
C
$$\left[{\mathrm\mu}_\mathrm o\right]\;=\;\mathrm{MLT}^{-2}\mathrm I^{-2}$$
D
$$\left[{\mathrm\mu}_\mathrm o\right]\;=\;\mathrm{ML}^2\mathrm T^{-1}\;\mathrm I$$

Explanation

From coulombs law, we know that

$$F = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {{r^2}}} \Rightarrow {\varepsilon _0} = {1 \over {4\pi }}{{{q_1}{q_2}} \over {{r^2}F}}$$

Therefore, $${\varepsilon _0} = {{[{I^2}{T^2}]} \over {[{L^2}][ML{T^{ - 2}}]}} = {M^{ - 1}}{L^{ - 3}}{T^4}{I^2}$$

Now, $$c = {1 \over {\sqrt {{\varepsilon _0}{\mu _0}} }} \Rightarrow {c^2} = {1 \over {{\varepsilon _0}{\mu _0}}}$$

Therefore, $${\mu _0} = {1 \over {{c^2}{\varepsilon _0}}} = {1 \over {{{[L{T^{ - 1}}]}^2}[{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}]}} = [ML{T^{ - 2}}{I^{ - 2}}]$$

Answer (B), (C)
3

IIT-JEE 1998

MCQ (More than One Correct Answer)
The SI unit of inductance, the henry can be written as
A
weber/ampere
B
volt-sec/amp
C
Joule/(ampere)2
D
ohm-second

Explanation

Induction $$L = {\phi \over i} = $$ weber/Ampere

Also, $$e = - L\left( {{{di} \over {dt}}} \right) \Rightarrow L = {{ - e} \over {(di/dt)}} = {{volt - {\mathop{\rm second}\nolimits} } \over {Ampere}}$$

Also, $$U = {1 \over 2}L{i^2} \Rightarrow L = {{2U} \over {{i^2}}} = {{Joule} \over {{{(Ampere)}^2}}}$$

and $$U = {1 \over 2}L{i^2} = {i^2}RT \Rightarrow L = RT = $$ Ohm-second

Answer (A), (B), (C), (D)
4

IIT-JEE 1995 Screening

MCQ (More than One Correct Answer)
The pairs of physical quantities that have the same dimensions is (are):
A
Reynolds number and coefficient of friction
B
Curie and frequency of a light wave
C
Latent heat and gravitational potential
D
Planck's constant and torque

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