A metal box is dragged on the ground through a distance of 42m. Coefficient of friction between the box and the ground is 0.2. Then the rise in temperature of the box if specific heat of material of the box is 0.98 C.G.S units and J = 4.2 J/cal

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

0.01°C

0.02°C

0.04°C

0.05°C

answer is B.

(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

Work done = Heat developed
$42 \times 0 .2 \times 9 .8 \times m = m \times 0 .98 \times 4200 ΔT ΔT = \frac{42 \times 0 .2 \times 9 .8}{4200 \times 0 .98} = 0 {.02}^{\circ} C$

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 Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE