JEE Main 2023 (Online) 1st February Evening Shift | Units & Measurements Question 102 | Physics

1

JEE Main 2023 (Online) 1st February Evening Shift

MCQ (Single Correct Answer)

+4

-1

If the velocity of light $$\mathrm{c}$$, universal gravitational constant $$\mathrm{G}$$ and Planck's constant $$\mathrm{h}$$ are chosen as fundamental quantities. The dimensions of mass in the new system is :

A

$$\left[\mathrm{h}^{1} \mathrm{c}^{1} \mathrm{G}^{-1}\right]$$

B

$$\left[\mathrm{h}^{-1 / 2} \mathrm{c}^{1 / 2} \mathrm{G}^{1 / 2}\right]$$

C

$$\left[\mathrm{h}^{1 / 2} \mathrm{c}^{1 / 2} \mathrm{G}^{-1 / 2}\right]$$

D

$$\left[\mathrm{h}^{1 / 2} \mathrm{c}^{-1 / 2} \mathrm{G}^{1}\right]$$

2

JEE Main 2023 (Online) 31st January Evening Shift

MCQ (Single Correct Answer)

+4

-1

Match List I with List II

LIST I		LIST II
A.	Angular momentum	I.	$\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]$
B.	Torque	II.	$\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right]$
C.	Stress	III	$\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]$
D.	Pressure gradient	IV.	$\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$

Choose the correct answer from the options given below:

A

A - I, B - IV, C - III, D - II

B

A - III, B - I, C - IV, D - II

C

A - IV, B - II, C - I, D - III

D

A - II, B - III, C - IV, D - I

3

JEE Main 2023 (Online) 30th January Evening Shift

MCQ (Single Correct Answer)

+4

-1

Match List I with List II:

List I		List II
A.	Torque	I.	$\mathrm{kg} \mathrm{m}^{-1} \mathrm{~s}^{-2}$
B.	Energy density	II.	$\mathrm{kg} \,\mathrm{ms}^{-1}$
C.	Pressure gradient	III.	$\mathrm{kg}\, \mathrm{m}^{-2} \mathrm{~s}^{-2}$
D.	Impulse	IV.	$\mathrm{kg} \,\mathrm{m}^{2} \mathrm{~s}^{-2}$

Choose the correct answer from the options given below:

A

A-IV, B-I, C-III, D-II

B

A-IV, B-I, C-II, D-III

C

A-I, B-IV, C-III, D-II

D

A-IV, B-III, C-I, D-II

4

JEE Main 2023 (Online) 29th January Evening Shift

MCQ (Single Correct Answer)

+4

-1

The equation of a circle is given by $$x^2+y^2=a^2$$, 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 : $${(x - At)^2} + {\left( {y - {t \over B}} \right)^2} = {a^2}$$. The dimensions of t is given as $$[\mathrm{T^{-1}]}$$.

A

$$\mathrm{A=[L^{-1}T^{-1}],B=[LT^{-1}]}$$

B

$$\mathrm{A=[L^{-1}T^{-1}],B=[LT]}$$

C

$$\mathrm{A=[LT],B=[L^{-1}T^{-1}]}$$

D

$$\mathrm{A=[L^{-1}T],B=[LT^{-1}]}$$

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