JEE Advanced 2015 Paper 1 Offline | Units & Measurements Question 20 | Physics

JEE Advanced 2015 Paper 1 Offline

MCQ (More than One Correct Answer)

-2

Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:

If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.

If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.

If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.

If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.

JEE Advanced 2015 Paper 1 Offline

MCQ (More than One Correct Answer)

-2

Planck's constant h, speed of light c and gravitational constant G are used to form a unit of length L and a unit of mass M. Then the correct option(s) is(are)

$$M \propto \sqrt c $$

$$M \propto \sqrt G $$

$$L \propto \sqrt h $$

$$L \propto \sqrt G $$

IIT-JEE 2010 Paper 1 Offline

MCQ (More than One Correct Answer)

-1

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?

Error ΔT in measuring T, the time period, is 0.05 seconds

Error ΔT in measuring T, the time period, is 1 second

Percentage error in the determination of g is 5%

Percentage error in the determination of g is 2.5%

IIT-JEE 1998

MCQ (More than One Correct Answer)

-0.5

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,

$$\left[{\mathrm\varepsilon}_\mathrm o\right]\;=\;\mathrm M^{-1}\mathrm L^{-3}\mathrm T^2\;\mathrm I$$

$$\left[{\mathrm\varepsilon}_\mathrm o\right]\;=\;\mathrm M^{-1}\mathrm L^{-3}\mathrm T^4\mathrm I^2$$

$$\left[{\mathrm\mu}_\mathrm o\right]\;=\;\mathrm{MLT}^{-2}\mathrm I^{-2}$$

$$\left[{\mathrm\mu}_\mathrm o\right]\;=\;\mathrm{ML}^2\mathrm T^{-1}\;\mathrm I$$

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