Question

Two identical balls A and B are moving with same velocity. If velocity of A is reduced to half and of B to zero, then the rise in temperatures of
A to that of B is
(A) 3:4
(B) 4:1
(C) 2.1

Answer 1

correct answer is option : 1 = 3: 4

Explanation:

For ball A:

mass= m

velocity=v

specific  heat =s

Energy dissipated for ball A= Initial k.e- final k.e

=

$\frac{mv^{2} }{2}</p><p>m(v/2)^{2} / 2$

=3/8 mv²

msΔTA= 3/8 mv²

For ball b:

Energy dissipated by ball b= 1/2 m v² - 0

=1/2 m v²

msΔTb=1/2 m v²

ΔTb= v²/2s

For ball b:

Energy dissipated by ball b= 1/2 m v² - 0

=1/2 m v²

msΔTb=1/2 m v²

ΔTb= v²/2s

ΔTa/ Δtb= 3v²x2s/ 8s x v²

= 3/4

∴ΔTa:ΔTb= 3:4