Q: Three rods made of same material and having same cross-section have been joined as shown in figure. Each rod is of same length. The left and right ends are kept at 0^{\circ}C and 90^{\circ}C respectively. The temperature of the junction of the three rods will be

Let \theta ^{\circ}C be the temperature at B . Let Q be the heat flowing per second from A to B on account of temperature difference by conductivity. \therefore Q = \dfrac {KA(90 - \theta)}{l} ... (i) where k = thermal conductivity of the rod, A = Area of cross section of the rod, l = length of the rod. By symmetry, the same will be the case for heat flow from C to B . \therefore The heat flowing per second from B to D will be 2Q = \dfrac {KA(\theta - 0)}{l} .... (ii) Dividing Eq. (ii) by Eq. (i) 2 = \dfrac {\theta}{90 - \theta} \Rightarrow \theta = 60^{\circ}C .

Question 1

A circular metal loop is heated at point  O  as shown in above Figure.i) In which direction would heat flow in the loop?ii) In which order the pins at points P, Q and R fixed with the help of wax fall if points O, P, Q and R are equidistant from each other?

Accepted Answer

According to the given condition.i) The heat will flow in the direction from O to P and O to R.ii) Firstly the pins at R and P will fail concurrently followed by the pin at Q.

Question 2

The temperature of equal masses of three different liquids  A, B  and  C  are  12^{\circ}C, 19^{\circ}C  and  28^{\circ}C  respectively. The temperature when  A  and  B  are mixed is  16^{\circ}C  and when  B  and  C  are mixed it is  23^{\circ}C . What should be the temperature when  A  and  C  are mixed?

Accepted Answer

Let  m  be the mass of each liquid and  S_{A}, S_{B}, S_{C}  be specific heats of liquids  A, B  and  C  respectively. When  A  and  B  are mixed. The final temperature is  16^{\circ}C . 	herefore  Heat gained by  A =  heat lost by  B i.e.,  mS_{A} = (16 - 12) = mS_{B} (19 - 16) i.e.,  S_{B} = \dfrac {4}{3}S_{A} ..... (i) When  B  and  C  are mixed. Heat gained by  B =  Heat lost by  C i.e.,  mS_{B} = (23 - 19) = mS_{C} (28 - 23) i.e.,  S_{C} = \dfrac {4}{5}S_{B} ..... (ii) From eq. (i) and (ii) S_{C} = \dfrac {4}{5}	imes \dfrac {4}{3} S_{A} = \dfrac {16}{15}S_{A} When  A  and  C  are mixed, let the final temperature be  	heta 	herefore mS_{A} (	heta - 12) = mS_{C} (28 - 	heta) i.e.,  	heta - 12 = \dfrac {16}{15} (28 - 	heta) By solving, we get, 	heta = \dfrac {628}{31} = 20.26^{\circ}C .

Question 3

A black body at a temperature of  227^{\circ}C  radiates heat energy at the rate  5\;cal/cm^2/s . At a temperature of  727^{\circ}C , the rate of heat radiated per unit area in  cal/cm^2  will be:

Accepted Answer

Heat energy per unit time           R = \sigma AeT^4 Thus heat energy per unit area        R'  = \sigma e T^4 \implies      \dfrac{R'_2}{R'_1}  = \bigg( \dfrac{T_2}{T_1} \bigg)^4 Initial temperature of the body        T_1   =227 ^oC   = (227+ 273) K  = 500  K Final temperature of the body        T_2   =727 ^oC   = (727+ 273) K  = 1000  K 	herefore       \dfrac{R'_2}{5}  = \bigg( \dfrac{1000}{500} \bigg)^4 \dfrac{R'_2}{5}  = 16                 \implies   R'_2  = 80     cal/cm^2

Question 4

A body cools from  {80}^{\circ} C to  {50}^{\circ} C in  5  minutes. Calculate the time it takes to cool from  {60}^{\circ} C to  {30}^{\circ} C. The temperature of the surroundings is  {20}^{\circ} C.

Accepted Answer

Here, Initial temperature  ( {T}_{i}) = {80}^{\circ} C Final temperature  ( {T}_{f}) = {50}^{\circ} C Temperature of the surrounding  ( {T}_{0}) = {20}^{\circ} C   t = 5  min According to Newton's law of cooling,Rate of cooling  \dfrac{dT}{dt} = K[\dfrac{\left({T}_{i}+{T}_{f}ight)}{2}-{T}_{o}] \dfrac{\left({T}_{f}-{T}_{i}ight)}{t}=K\left[\dfrac{\left(80 +50ight)}{2} - 20ight] \dfrac{80-50}{5} = K[ 65 - 20] 6 = K	imes  45 K = \dfrac{6}{45} = \dfrac{2}{15}  In second condition, initial temperature  ={T}_{i}= {60}^{\circ} C Final temperature  {T}_{f}={30}^{\circ} C Time taken for cooling is  t  According to Newton's law of cooling  \dfrac{( 60 - 30)}{t} = \dfrac{2}{15} \left[ \dfrac{\left(60+30ight)}{2} -20ight] \dfrac{30}{t}=\dfrac{2}{15}	imes  25 \dfrac{30}{t} = \dfrac{50}{15}=\dfrac{10}{3} 	herefore t = 9  min

Question 5

The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.

Accepted Answer

The conversion formulae are  t_{c}=t_{k}-273.15 t_{f}=\dfrac{9}{5}t_{c}+32 Substituting the given values, we get  	ext{Neon:}\ 24.57\ K=-248.43\ ^o C = -415.44\ ^o F 	ext{CO:}\ 216.55\ K=-56.45\ ^o C=-69.8\ ^oF

Question 6

At what temperature, the Celsius and Fahrenheit scales will read the same ?

Accepted Answer

The formulas for converting between degree Celsius and degree Fahrenheit are: °F = (°C * 9/5) + 32 °C = (°F - 32) * 5/9 To find the temperature when both are equal, we use an old algebra trick and just set ºF = ºC and solve one of the equations. °C = (°C * 9/5) + 32 °C - (°C * 9/5) = 32 -4/5 * °C = 32 °C = -32 * 5/4 °C = -40 °F = (°F * 9/5) + 32 °F - (°F * 9/5) = 32 -4/5 * °F = 32 °F = -32 * 5/4 °F = -40 So the temperature when both the Celsius and Fahrenheit scales are the same is  -40 degrees.

Question 7

Six identical conducting rods are joined as shown in figure. Points  A  and  D  are maintained at temperature of  200 ^\circ C  and  20 ^\circ C  respectively. The temperature of junction  B  will be :

Accepted Answer

{ k }_{ BC }=k  (wheat stone bridge)  \dfrac { { k }A(200-{ t }_{ B }) }{ l } =\dfrac { kA({ t }_{ B }-{ t }_{ C }) }{ l } =\dfrac { kA({ t }_{ C }-20) }{ l }   { t }_{ B }={ 140 }^{ \circ }C{ \quad \& \quad t }_{ C }={ 80 }^{ \circ }C

Question 8

A hole is drilled in a copper sheet.  The diameter of the hole is 4.24 cm at  27.0 ^oC . What is the change in the diameter of the hole when the sheet is heated to  227^oC  ? Coefficient  of the linear expansion of copper  =1.70	imes 10^{-5}K^{-1}

Accepted Answer

Initial temperature,  T_1=27.0^oC Diameter of the hole at  T_1, d_1=4.24cm Final temperature,  T_2=227^oC Diameter of the hole at  T_2=D_2 Co-efficient of linear expansion of copper,  \alpha_{Cu} =1.70	imes 10^{-5}K^{-1} For co-efficient of superficial expansion β,and change in temperature  	riangle T , we have the relation:Change in area  (	riangle)  /  Original area (A)  =  \beta 	riangle T [(\pi d_2^2/4)-(\pi d_2^2/4)]/(\pi d_1^1/4)=	riangle A/A 	herefore 	riangle A/A = (d_2^2-d_1^2)/d_1^2 But  \beta = 2\alpha 	herefore (d_2^2-d_1^2)/d_1^2=2\alpha	riangle T (d_2^2/d_1^2)-1=2\alpha(T_2-T_1) d_2^2/4.24^2=2	imes 1.7	imes 10^{-5}(227-27)+1 d_2^2=17.98	imes 1.0068=18.1 	herefore d_2=4.2544 cm Change in diameter  =d_2-d_1=4.2544-4.24=0.0144 cm Hence, the diameter increases by  1.44	imes 10^{-2}cm

Question 9

Three rods made of same material and having same cross-section have been joined as shown in figure. Each rod is of same length. The left and right ends are kept at  0^{\circ}C  and  90^{\circ}C  respectively. The temperature of the junction of the three rods will be

Accepted Answer

Let  	heta ^{\circ}C  be the temperature at  B . Let  Q  be the heat flowing per second from  A  to  B  on account of temperature difference by conductivity. 	herefore Q = \dfrac {KA(90 - 	heta)}{l} ... (i) where  k =  thermal conductivity of the rod,  A =  Area of cross section of the rod,  l =  length of the rod. By symmetry, the same will be the case for heat flow from  C  to  B . 	herefore  The heat flowing per second from  B  to  D  will be 2Q = \dfrac {KA(	heta - 0)}{l} .... (ii) Dividing Eq. (ii) by Eq. (i) 2 = \dfrac {	heta}{90 - 	heta} \Rightarrow 	heta = 60^{\circ}C .

Question 10

Two rods of length  l_1  and  l_2  are made of materials whose coefficients of linear expansion are  \alpha_1  and  \alpha_2 . If the difference between their lengths is independent of temperature, then :

Accepted Answer

Length of Rods  =L_{1}  &  L_{2} Coefficient of thermal Expansion  =\alpha_{1}  &  \alpha_{2} According to the question 	riangle L_{1}=	riangle L_{2} L_{1}\alpha_{1}	riangle T=L_{2}\alpha_{2}	riangle T \dfrac{L_{1}}{L_{2}}=\dfrac{\alpha_{2}}{\alpha_{1}}