Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

A block of ice of mass 50 kg is sliding on a horizontal plane. It starts with speed 5 ms-1 and stops after moving through some distance. The mass of ice that has melted due to friction between the block and the surface is (Assuming that no energy is lost and latent heat of fusion of ice is 80 cal g -1 , J = 4.2 J cal-1)

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

a

2.86 g

b

3.86 g

c

0.86 g

d

1.86 g

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Given,   m=50kg,v=5ms−1,L=80calg−1J=42Jcal−1 and m=?Heat lost, i.e. Q1 = work done = K (kinetic energy)=12mv2=12×50×52=625Jand heat gained, i.e.   Q2=m'×L=80×m'×4.2∵   Q1  =  Q2625=m'×8×42⇒m'=1.86g
Watch 3-min video & get full concept clarity

courses

No courses found

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon