The weight of a person is 60 kg If he gets 105 calories of heat through food and the efficiency of his body is 28 %, then upto how much height he can climb? (Take .g=10 m s -2)

Question

The weight of a person is 60 kg If he gets 105 calories of heat through food and the efficiency of his body is 28 %, then upto how much height he can climb? (Take .g=10 m s -2) (A) 100 m (B) 196 m (C) 400 m (D) 1000 m

Tardigrade Admin · Accepted Answer

W=η ⋅ Q=((28/100)) ⋅(105. cal ) W=((28/100) × 105 × 4.2) J But W=m ⋅ g ⋅ h h=(W/m g)=((28 × 105 × 4.2/100 × 60 × 10))=196 m.

Q. The weight of a person is 60 kg If he gets $1 0^{5}$ calories of heat through food and the efficiency of his body is $28%,$ then upto how much height he can climb? (Take $g = 10 m s^{- 2})$

100 m

196 m

400 m

1000 m

$W = η \cdot Q = (\frac{28}{100}) \cdot (1 0^{5}$ cal $)$
$W = (\frac{28}{100} \times 1 0^{5} \times 4.2) J$
But $W = m \cdot g \cdot h$
$h = \frac{W}{m g} = (\frac{28 \times 1 0 ^{5} \times 4.2}{100 \times 60 \times 10}) = 196 m$ .

Q. The weight of a person is 60 kg If he gets 105 calories of heat through food and the efficiency of his body is 28%, then upto how much height he can climb? (Take g=10ms−2)