Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Eight Laws of Motion Physics Physics Laws of Motion

Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is $0.15 (g = 10 {m s}^{- 2})$ .

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mn>50</mn> <msup> <mrow> <mrow> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">s</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mn>1.2</mn> <msup> <mrow> <mrow> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">s</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mn>150</mn> <msup> <mrow> <mrow> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">s</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mn>1.5</mn> <msup> <mrow> <mrow> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">s</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> </math> </span></p>

11 Views|Posted 5 months ago

Asked by Shiksha User

1 Answer

Sort by:

A

Answered by

Alok Kumar Singh | Contributor-Level 10

5 months ago

Correct Option - 4

Detailed Solution:

$R = R_{0} (1 + α Δ T)$

$\begin{array}{r} \Rightarrow 6.8 = 2 [1 + α * (80 - 0)] \\ \Rightarrow α = \frac{3.4 - 1}{80} \\ = 0.03 \\ = 3 * 10^{- 2}^{?} C^{- 1} \end{array}$

Upvote

Thumbs Down Icon

Similar Questions for you

Two blocks of mass 10 kg and 5 kg are connected with a light string which passing over an ideal pulley attached with a fixed smooth wedge. If the system is released from rest, then the acceleration of the blocks will be

Alok Kumar Singh

A block rests on an inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between block and the plane is 0.8. If the frictional force on the block is 10 N, then mass of block (in kg) (Take g = 10 m/s² )

Alok Kumar Singh

Bullets of 0.03 kg mass each hit a plate at a rate of 200 bullets per second with velocity of 50 m/s and reflect back with velocity of 30 m/s. The average force acting on the plate in newton is

Alok Kumar Singh

The root mean square speed of molecules of a gas is 1260 m/s. The average speed of the gas molecules is

Alok Kumar Singh

= 0.92 * 1260 = 1161 m/s

Find the acceleration of 2 kg block shown in the diagram. (neglect friction)

Alok Kumar Singh

For 2 kg block

T – 2g sin37 = 2a . (i)

For 4 kg block

4g – 2T = $\frac{4 a}{2}$

2g – T = a . (ii)

T = (2g – a)

2g – a – 2g × $\frac{3}{5}$ = 2a

3a = 2g × $\frac{2}{5}$

$a = \frac{4 g}{15}$

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K

Colleges

|

1.2K

Exams

|

6.9L

Reviews

|

1.8M

Answers

Learn more about...

Physics NCERT Exemplar Solutions Class 11th Chapter Eight 2025

Physics NCERT Exemplar Solutions Class 11th Chapter Eight 2025

View Exam Details

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

or

See what others like you are asking & answering