(A) The dimensional formulas are calculated as follows:$(1)$ Torque $(\tau) = \text{Force} \times \text{Distance} = [MLT^{-2}] \times [L] = [ML^2T^{-2

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The dimensional formulas are calculated as follows:$(1)$ Torque $(\tau) = \text{Force} \times \text{Distance} = [MLT^{-2}] \times [L] = [ML^2T^{-2}]$. Thus, $(1) - (c)$.$(2)$ Angular momentum $(L) = \text{Moment of inertia} \times \text{Angular velocity} = [ML^2] \times [T^{-1}] = [ML^2T^{-1}]$. Thus, $(2) - (b)$.$(3)$ Linear momentum $(p) = \text{Mass} \times \text{Velocity} = [M] \times [LT^{-1}] = [MLT^{-1}]$. Thus, $(3) - (a)$.Therefore, the correct matching is $1-c, 2-b, 3-a$.

Class 11 Physics Units, Dimensions and Measurement Dimensions and Dimensional Formula

Medium

Match the physical quantities in Column-$I$ with their corresponding dimensional formulas in Column-$II$.

Column-$I$ Column-$II$

$(1)$ Torque $(a)$ $M^1L^1T^{-1}$

$(2)$ Angular momentum $(b)$ $M^1L^2T^{-1}$

$(3)$ Linear momentum $(c)$ $M^1L^2T^{-2}$

Column-$I$	Column-$II$
$(1)$ Torque	$(a)$ $M^1L^1T^{-1}$
$(2)$ Angular momentum	$(b)$ $M^1L^2T^{-1}$
$(3)$ Linear momentum	$(c)$ $M^1L^2T^{-2}$

A
$1-c, 2-b, 3-a$
B
$1-b, 2-c, 3-a$
C
$1-c, 2-a, 3-b$
D
$1-a, 2-b, 3-c$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Units, Dimensions and Measurement

Subtopic

Dimensions and Dimensional Formula

Write the dimensional formula for the intensity of radiation.

EasyGUJCET 2007

View Solution

Given below are two statements :
Statement $(I)$ : Dimensions of specific heat are $\left[L^2 \,T^{-2} \,K^{-1}\right]$.
Statement $(II)$ : Dimensions of gas constant are $\left[ML^2 \,T^{-1} \,K^{-1}\right]$.

DifficultJEE MAIN 2024

View Solution

Match List $I$ with List $II$:
List $I$ List $II$
$A$. Spring constant $I$. $(T^{-1})$
$B$. Angular speed $II$. $(MT^{-2})$
$C$. Angular momentum $III$. $(ML^2)$
$D$. Moment of inertia $IV$. $(ML^2T^{-1})$

Choose the correct answer from the options given below:

MediumJEE MAIN 2023

View Solution

The position of a particle at time $t$ is given by the equation $x(t) = \frac{v_0}{A}(1 - e^{-At})$, where $v_0$ is a constant and $A > 0$. The dimensions of $v_0$ and $A$ respectively are:

MediumTS EAMCET 2004

View Solution

Which of the following quantities is dimensionless?

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

List $I$	List $II$
$A$. Spring constant	$I$. $(T^{-1})$
$B$. Angular speed	$II$. $(MT^{-2})$
$C$. Angular momentum	$III$. $(ML^2)$
$D$. Moment of inertia	$IV$. $(ML^2T^{-1})$

Match the physical quantities in Column-$I$ with their corresponding dimensional formulas in Column-$II$. Column-$I$ Column-$II$ $(1)$ Torque $(a)$ $M^1L^1T^{-1}$ $(2)$ Angular momentum $(b)$ $M^1L^2T^{-1}$ $(3)$ Linear momentum $(c)$ $M^1L^2T^{-2}$

Similar Questions

Write the dimensional formula for the intensity of radiation.

Given below are two statements :Statement $(I)$ : Dimensions of specific heat are $\left[L^2 \,T^{-2} \,K^{-1}\right]$.Statement $(II)$ : Dimensions of gas constant are $\left[ML^2 \,T^{-1} \,K^{-1}\right]$.

Match List $I$ with List $II$:List $I$List $II$$A$. Spring constant$I$. $(T^{-1})$$B$. Angular speed$II$. $(MT^{-2})$$C$. Angular momentum$III$. $(ML^2)$$D$. Moment of inertia$IV$. $(ML^2T^{-1})$Choose the correct answer from the options given below:

The position of a particle at time $t$ is given by the equation $x(t) = \frac{v_0}{A}(1 - e^{-At})$, where $v_0$ is a constant and $A > 0$. The dimensions of $v_0$ and $A$ respectively are:

Which of the following quantities is dimensionless?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Match the physical quantities in Column-$I$ with their corresponding dimensional formulas in Column-$II$.

Column-$I$ Column-$II$

$(1)$ Torque $(a)$ $M^1L^1T^{-1}$

$(2)$ Angular momentum $(b)$ $M^1L^2T^{-1}$

$(3)$ Linear momentum $(c)$ $M^1L^2T^{-2}$

Given below are two statements :
Statement $(I)$ : Dimensions of specific heat are $\left[L^2 \,T^{-2} \,K^{-1}\right]$.
Statement $(II)$ : Dimensions of gas constant are $\left[ML^2 \,T^{-1} \,K^{-1}\right]$.

Match List $I$ with List $II$:
List $I$ List $II$
$A$. Spring constant $I$. $(T^{-1})$
$B$. Angular speed $II$. $(MT^{-2})$
$C$. Angular momentum $III$. $(ML^2)$
$D$. Moment of inertia $IV$. $(ML^2T^{-1})$

Choose the correct answer from the options given below: