The coefficient of apparent expansion of mercury in glass is 16400-0C- A glass bulb at 0 0C is 260 g- The glass bulb is heated to 100 0C- Mass of the mercury over flown will be -

The correct option is D #160; 4gm γ_a = m_2 m_1m_1 Δ T=m_2 m_1260 × (100 0) m_2 m_1 = 16400× 260 × 100 = 26064 = 4.06 gm ...

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Standard X

Physics

Expansion In Liquids and Gases

The coefficie...

Question

The coefficient of apparent expansion of mercury in glass is $\frac{1}{6400} /^{0} C$ . A glass bulb at 0 $^{0}$ C is 260 g. The glass bulb is heated to 100 $^{0}$ C. Mass of the mercury over flown will be :

\frac{1}{2} g m

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\frac{1}{4} g m

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\frac{1}{3} g m

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4 g m

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Solution

The correct option is D $4 g m$
$γ_{a} = \frac{m_{2} - m_{1}}{m_{1} Δ T} = \frac{m_{2} - m_{1}}{260 \times (100 - 0)}$
$m_{2} - m_{1} = \frac{1}{6400} \times 260 \times 100 = \frac{260}{64} = 4.06 g m$

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The coefficient of apparent expansion of mercury in glass is 16400/0C. A glass bulb at 00C is 260 g. The glass bulb is heated to 1000C. Mass of the mercury over flown will be :

The correct option is D 4gmγa=m2−m1m1ΔT=m2−m1260×(100−0)m2−m1=16400×260×100=26064=4.06gm

The coefficient of apparent expansion of mercury in glass is $\frac{1}{6400} /^{0} C$ . A glass bulb at 0 $^{0}$ C is 260 g. The glass bulb is heated to 100 $^{0}$ C. Mass of the mercury over flown will be :

The correct option is D $4 g m$
$γ_{a} = \frac{m_{2} - m_{1}}{m_{1} Δ T} = \frac{m_{2} - m_{1}}{260 \times (100 - 0)}$
$m_{2} - m_{1} = \frac{1}{6400} \times 260 \times 100 = \frac{260}{64} = 4.06 g m$