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AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper
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© examsnet.com
Question : 98
Total: 160
A hammer of mass 200 kg strikes a steel block of mass 200 g with a velocity 8
m
s
−
1
. If 23% of the energy is utilized to heat the steel block, the rise in temperature of the block is (specific heat capacity of steel = 460 J
k
g
−
1
K
−
1
)
8 °C
16 °C
12 °C
24 °C
Validate
Solution:
The expression for the kinetic energy is given by,
K
E
=
1
2
m
v
2
Substitute the given values in the above equation.
K
E
=
1
2
×
200
×
8
2
=
6400
J
According to the statement of the question,
H
=
6400
×
23
100
=
1472
J
The rise in temperature of the steel is calculated as,
∆
T
=
H
m
s
=
1472
460
×
0.2
=
16
K
© examsnet.com
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