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Thermal Properties of Matter Test - 60

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Thermal Properties of Matter Test - 60
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  • Question 1
    1 / -0
    In cold countries, water pipes sometimes burst, because
    Solution
    In anomalous expansion, water contracts on heating and expands on cooling in the range $$0^\circ C$$ to $$4^\circ C$$. Therefore water pipes sometimes burst, in cold countries.
  • Question 2
    1 / -0
    Wires A and B have identical lengths and have circular cross-sections. The radius of A is twice the radius of B i.e. $$r_A=2r_B$$. For a given temperature difference between the two ends, both wires conduct heat at the same rate. The relation between the thermal conductivities is given by :
    Solution
    $$\dfrac{Q}{t}=\dfrac{KA\Delta \theta}{l}\,\rightarrow \dfrac{K_A}{K_B}=\dfrac{A_B}{A_A}=\Bigg(\dfrac{r_B}{r_B}\Bigg)^2=\dfrac{1}{4}\,\implies K_A=\dfrac{K_B}{4}$$
  • Question 3
    1 / -0
    The length of the two rods made up of the same metal and having the same area of the cross-section are $$0.6$$ m and $$0.8$$ m respectively. The temperature between the ends of the first rod is $$90 ^\circ C$$ and $$60 ^\circ C$$ and that for the other rod is $$150$$ and $$110 ^\circ C$$. For which rod the rate of conduction will be greater
    Solution
    $$\dfrac{Q_1}{t}=\dfrac{KA(90-60)}{0.6}=50 \ KA$$ and $$\dfrac{Q_2}{t}=\dfrac{KA(150-110)}{0.8}=50 \ KA$$ .
  • Question 4
    1 / -0
    The coefficients of thermal conductivity of copper, mercury, and glass are respectively K, K, and K such that $$K > K > K$$. If the same quantity of heat is to flow per second per unit area of each and corresponding temperature gradients are X, X, and X, then 
    Solution
    The heat flowing is given by $$\dfrac{Q}{At}=K\dfrac{\Delta \theta}{l}\,\implies K\dfrac{\Delta \theta}{l}=constant\,\implies \dfrac{\Delta \theta}{l} \propto \dfrac{1}{K}$$
    Hence If $$K_c > K_m > K_g$$, then 
    $$\Bigg(\dfrac{\Delta \theta }{l}\Bigg)_c < \Bigg(\dfrac{\Delta \theta }{l}\Bigg)_m < \Bigg(\dfrac{\Delta \theta }{l}\Bigg)_g\,\implies X_c < X_m < X_g$$
    because higher K implies lower value of the temperature gradient. 
  • Question 5
    1 / -0
    One likes to sit under sunshine in the winter season, because
    Solution
    Heat flows from hot air to cold body so the person feels comfortable that's why one likes to sit under sunshine in the winter season.
  • Question 6
    1 / -0
    Calorimeters are made of which of the following
    Solution
    Calorimeters are made by conducting materials, mainly metals
  • Question 7
    1 / -0
    The layers of the atmosphere are heated through 
    Solution
    The layers if the atmosphere are heated through Radiation from sun and convection currents, are responsible for heating.
  • Question 8
    1 / -0
    Two rods of the same length and cross-section are joined along the length. Thermal conductivities of the first and second rod are $$K_1$$ and $$K_2$$. The temperature of the free ends of the first and second rods are maintained at $$\theta_1$$ and $$\theta_2$$ respectively. The temperature of the common junction is :
    Solution
    At steady-state, the rate of heat flow for both blocks will be the same :
    i.e.$$ \dfrac{K_1A(\theta_1-\theta)}{l_1}= \dfrac{K_2A(\theta-\theta_2)}{l_2}\,(given\,l_1=l_2)$$
    $$\Rightarrow K_1A(\theta_1-\theta)= K_2A(\theta-\theta_2)\,\Rightarrow \theta=\dfrac{K_1\theta_1+K_2\theta_2}{K_1+K_2}$$

  • Question 9
    1 / -0
    A body of length $$1m$$ having a cross-sectional area $$0.75m^2$$ has heat flow through it at the rate of $$6000 \ Joule/sec$$. Then find the temperature difference if $$K=200Jm^{-1}K^{-1}$$
    Solution
    $$\dfrac{Q}{t}=\dfrac{KA \Delta \theta}{l}\,\Rightarrow 6000=\dfrac{200 \times 0.75 \times  \Delta \theta}{1}$$ .
    $$\therefore \Delta \theta=\dfrac{6000 \times 1}{2000 \times 0.75}=40 ^\circ C$$
  • Question 10
    1 / -0
    Two cylinders P and Q have the same length and diameter and are made of different materials having thermal conductivities in the ratio $$2: 3$$. These two cylinders are combined to make a cylinder. One end of P is kept at $$100 ^\circ C$$ and another end of Q at $$0 ^\circ C$$. The temperature at the interface of P and Q is 
    Solution
    Temperature of interface $$\theta=\dfrac{K_1\theta_1+K_1\theta_1}{K_1+K_2} $$ ,
    where $$K_1 $$ = $$2K$$ and $$K_2 $$ = $$3K$$  $$\Bigg(\therefore \dfrac{K_1}{K_2}=\dfrac{2}{3}\Bigg)$$

    $$\Rightarrow \theta=\dfrac{2K \times 100+3K \times 0}{2K+3K}=\dfrac{200 \ K}{5 \ K}=40 ^\circ C$$
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