. At $4^{\circ} C, 0.98$ of the volume of a body is immersed in water. The temperature at which the entire immersed in water is (neglect the expansion of the body) $(γ_{w} = 3.3 \times 10^{- 4} K^{- 1}) : -$

{40.8}^{\circ} C

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64 : 6^{\circ} C

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{60.6}^{\circ} C

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{58.8}^{\circ} C

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Solution

The correct option is C ${60.6}^{\circ} C$
$γ_{ω} = 3.3 \times 10^{\frac{- 4}{k}}$
for $1^{o} K \to 3.3 \times 10^{- 4}$ unit volume in creasen now for complete immerson of liquid we need to raise volume of liquid by $0.01$
So, $T = \frac{0.02}{3.3 \times 10^{- 4}}$
$T = 60.6 C$

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. At 4∘C,0.98 of the volume of a body is immersed in water. The temperature at which the entire immersed in water is (neglect the expansion of the body) (γw=3.3×10−4K−1):−

The correct option is C 60.6∘Cγω=3.3×10−4kfor 1oK→3.3×10−4 unit volume in creasen now for complete immerson of liquid we need to raise volume of liquid by 0.01So, T=0.023.3×10−4T=60.6C

. At $4^{\circ} C, 0.98$ of the volume of a body is immersed in water. The temperature at which the entire immersed in water is (neglect the expansion of the body) $(γ_{w} = 3.3 \times 10^{- 4} K^{- 1}) : -$

The correct option is C ${60.6}^{\circ} C$
$γ_{ω} = 3.3 \times 10^{\frac{- 4}{k}}$
for $1^{o} K \to 3.3 \times 10^{- 4}$ unit volume in creasen now for complete immerson of liquid we need to raise volume of liquid by $0.01$
So, $T = \frac{0.02}{3.3 \times 10^{- 4}}$
$T = 60.6 C$