A specific gravity bottle can hold 101 gm of a liquid at 0 $^{0}$ C and 100 gm at 100 $^{0}$ C. The co-efficient of apparent expansion of the liquid is :

1 \times 10^{- 3} /^{o} C

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1 \times 10^{- 6} /^{o} C

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1 \times 10^{- 5} /^{o} C

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1 \times 10^{- 4} /^{o} C

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Solution

The correct option is D $1 \times 10^{- 4} /^{o} C$
$m_{2} = m_{1} (1 + γ_{a} Δ t)$ ;

$∴ γ_{a} = \frac{m_{2} - m_{1}}{m_{1} Δ t}$ where $m_{1}$ and $m_{2}$ are mass of the liquids at $100^{0} C$ and $0^{0} C$ respectively.
$m_{2} = 101$
$m_{1} = 100$
$Δ t = 100 - 0 = 100$

$∴ γ_{a} = \frac{101 - 100}{100 \times 100} = 1 \times 10^{- 4}$

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A specific gravity bottle can hold 101 gm of a liquid at 00C and 100 gm at 1000C. The co-efficient of apparent expansion of the liquid is :

The correct option is D 1×10−4/oCm2=m1(1+γaΔt);∴γa=m2−m1m1Δt where m1 and m2 are mass of the liquids at 1000C and 00C respectively.m2=101m1=100Δt=100−0=100∴γa=101−100100×100=1×10−4

A specific gravity bottle can hold 101 gm of a liquid at 0 $^{0}$ C and 100 gm at 100 $^{0}$ C. The co-efficient of apparent expansion of the liquid is :