A 10 kg block is placed on top of a 40 kg block as shown in figure. A horizontal force F acting on B causes an acceleration of $2 m s^{- 2}$ to B. For this situation, mark out the correct statement(s).

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

The acceleration of A may be $2 m s^{- 2}$ or less than $2 m s^{- 2}$

The coefficient of friction between the blocks must be $0.2$ only.

The acceleration of A must also be $2 m s^{- 2}$

The coefficient of friction between the blocks may be $0.2$

answer is B, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} F = 10 a_{1} + 40 a_{2} \\ \Rightarrow 100 = 10 a_{1} + 40 \times 2 \\ \Rightarrow a_{1} = 2 m s^{- 2} \end{array}$

So acceleration of A must be 2ms-2 for given conditions to be satisfied.

$\begin{array}{l} f = 10 a_{1} = 10 \times 2 = 20 N \\ f \leq f_{l} \Rightarrow 20 \leq μ m_{A} g \\ \Rightarrow 20 \leq μ \times 10 g \Rightarrow μ \geq 0.2 \end{array}$

Hence, m can be greater or equal to $0.2$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!