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Test Index
JEE Main 2016 Paper
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Section:
Physics
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© examsnet.com
Question : 62 of 90
Marks:
+1
,
-0
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies
3.8
×
1
0
7
J
3.8\times10^{7}\mathrm{J}
3.8
×
1
0
7
J
of energy per kg which is converted to mechanical energy with a 20 % efficiency rate. Take
g
=
9.8
  
m
s
−
2
g= 9.8\;\mathrm{ms}^{-2}
g
=
9.8
ms
−
2
:
[Main 2016]
6.45
×
1
0
−
3
6.45\times10^{-3}
6.45
×
1
0
−
3
kg
9.89
×
1
0
−
3
9.89\times10^{-3}
9.89
×
1
0
−
3
kg
12.89
×
1
0
−
3
12.89\times10^{-3}
12.89
×
1
0
−
3
kg
2.45
×
1
0
−
3
2.45\times10^{-3}
2.45
×
1
0
−
3
kg
Validate
Solution:
Let fat used be ‘x’ kg
⇒ Mechanical energy available
=
x
×
3.8
×
1
0
7
×
20
100
=x\times3.8\times10^{7}\times\frac{20}{100}
=
x
×
3.8
×
1
0
7
×
100
20
​
⇒
x
×
3.8
×
1
0
7
×
20
100
=
9.8
×
1
0
4
\Rightarrow x\times3.8\times10^{7}\times\frac{20}{100}=9.8\times10^{4}
⇒
x
×
3.8
×
1
0
7
×
100
20
​
=
9.8
×
1
0
4
⇒
x
≈
12.89
×
1
0
−
3
\Rightarrow x\approx12.89\times10^{-3}
⇒
x
≈
12.89
×
1
0
−
3
kg
© examsnet.com
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