The density of water at 20^ C in 998 ~kg / m^3 and at a 40^ C 992 ~kg / m^3 . The cocfficient of volume expansion of water is:

Correct answer: A. 3 × 10^ -4 / ^ C According to the question aligned & T_1=20^ ^ C,T_2=40^ ^ C & _r=998 ~kg/ m^3 _ 40 =992 ~kg/ m^3 & So, _ T_2 = _ T_1 (1+v Δ T) = _ T_1 1+v(T_2-T_1) aligned Putting the values we get aligned & 992=(998)/(1+v(40-20)) & 992= 998 1+20 v & 992(1+20 v)=998 & 1+20 v=(998)/(992) & 20 v=(6)/(992) & v=(6)/(992) × (1)/(20)=3 ×10^ -4 / ^ ^ aligned

NEET 2013

Exam: NEET
Year: 2013
Subject: Physics
Chapter: Thermal Properties of Matter

NEET 2013 · Physics · Thermal Properties of Matter

The density of water at 20^ C in 998 ~kg / m^3 and at a 40^ C 992...

The density of water at

20^{\circ} \mathrm{C}

998 \mathrm{~kg} / \mathrm{m}^3

and at a

40^{\circ} \mathrm{C} 992 \mathrm{~kg} / \mathrm{m}^3

. The cocfficient of volume expansion of water is:

Options

A $3 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
B $2 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
C $6 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
D $10^{-3} \rho^{\circ} \mathrm{C}$

Correct answer

A. $3 \times 10^{-4} /{ }^{\circ} \mathrm{C}$

Step-by-step solution

According to the question

\begin{aligned}& T_1=20^{{}^{\circ}} \mathrm{C},T_2=40^{{}^{\circ}} \mathrm{C}\\ & \rho_r=998 \mathrm{~kg}/ \mathrm{m}^3 \rho_{40}=992 \mathrm{~kg}/ \mathrm{m}^3 \\ & \text{ So, } \rho_{\mathrm{T}_2}=\frac{\rho_{\mathrm{T}_1}}{(1+\mathrm{v} \Delta T)}=\frac{\rho_{\mathrm{T}_1}}{1+\mathrm{v}\left(T_2-T_1\right)} \end{aligned}

Putting the values we get

\begin{aligned}& 992=\frac{998}{1+v(40-20)}\\ & 992=\frac{998}{1+20 \mathrm{v}}\\ & \Rightarrow 992(1+20 \mathrm{v})=998 \\ & \Rightarrow \quad 1+20 \mathrm{v}=\frac{998}{992}\\ & \Rightarrow \quad 20 \mathrm{v}=\frac{6}{992}\\ & \Rightarrow \mathrm{v}=\frac{6}{992} \times \frac{1}{20}=3 \times10^{-4} /{ }^{{}^{\circ}} \end{aligned}

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