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Class 10 Class 12

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 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g=9.8 ms−2:

• 2.45 ×10−3 kg

• 6.45 x×10−3 kg

• 9.89 ×10−3 kg

• 12.89 ×10−3 kg

D.

12.89 ×10−3 kg

Given potential energy burnt by lifting weight

= mgh = 10 x 9.8 x 1 x 1000 = 9.8 x 104 J

If mass lost by a person be m, then energy dissipated

= m x 2 x 38 x 107 J /10

⇒ m = 5 x 10-3 x 9.8 / 3.8

= 12.89 x 10-3 kg

Given potential energy burnt by lifting weight

= mgh = 10 x 9.8 x 1 x 1000 = 9.8 x 104 J

If mass lost by a person be m, then energy dissipated

= m x 2 x 38 x 107 J /10

⇒ m = 5 x 10-3 x 9.8 / 3.8

= 12.89 x 10-3 kg

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What is watt?

SI unit of power is Watt.
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Define the unit joule.

Work done is said to be one joule if one newton of  force displaces the body through a distance of one meter in the direction of applied  force .
799 Views

Define watt.

Power is said to be one watt if one joule of work is done in one second.
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Is work a scalar or a vector quantity?

Work is dot product of two vectors. i.e. .

And dot product is a scalar quantity. Therefore, work is a scalar quantity.
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What are different units of energy?

The different units of energy are :

(i) Joule
(ii) Erg
(iii) eV
(iv) KWh
(v) Calorie.

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