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MHT CET 2021 Physics Solved Paper
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© examsnet.com
Question : 82
Total: 100
A lead bullet at
27
∘
C
melts when hit on a target. Assuming only
75
%
of heat is used to melt the bullet, the velocity of bullet at time of striking is
(Take, melting point of lead
=
327
∘
C
, specific heat of lead
=
0.03
c
a
l
∕
g
∘
C
, latent heat of fusion of lead
=
6
c
a
l
∕
g
)
330
m
∕
s
410
m
∕
s
470
m
∕
s
510
m
∕
s
Validate
Solution:
(b) Let mass of the bullet be
m
gram, then total heat required for bullet to just melt down
Q
1
=
m
c
∆
T
+
m
L
=
m
×
(
0.03
)
(
327
−
27
)
+
m
×
6
=
15
m
−
c
a
l
=
(
16
m
×
4
⋅
2
)
J
Now, when bullet is struck by obstacles, the loss in its mechanical energy
=
1
2
(
m
×
10
−
3
)
v
2
The energy absorbed by bullet,
Q
2
=
72
100
×
1
2
m
v
2
×
10
−
3
=
3
8
m
v
2
×
10
−
3
J
Now, the bullet will melt if
Q
2
≥
Q
1
i.e.,
3
8
m
v
2
×
10
−
3
≥
15
m
×
4.2
⇒
v
min
=
410
m
∕
s
© examsnet.com
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