Two rods, one made of aluminium and the other made of steel, having initial lengths \(l_1\) and \(l_2\) are connected together to form a single rod of length $l_1+l_2$. The coefficient of linear expansion for aluminium and steel are ${\alpha }_a$ and ${\alpha }_s$ respectively. If the length of each rod increases by the same amount when their temperature is raised by \(t^\circ \mathrm{C},\) then the ratio \(\frac{l_1}{l_1+l_2}\) is:
1. $\frac{{\alpha }_s}{{\alpha }_a}$

2. $\frac{{\alpha }_a}{{\alpha }_s}$

3. $\frac{{\alpha }_a}{{\alpha }_a+{\alpha }_s}$

4. $\frac{{\alpha }_s}{{\alpha }_a+{\alpha }_s}$

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

         
1. \(45^{\circ}\mathrm{C}\)
2. \(60^{\circ}\mathrm{C}\)
3. \(30^{\circ}\mathrm{C}\)
4. \(20^{\circ}\mathrm{C}\)$$

Hot coffee in a mug cools from \(90^{\circ}\mathrm{C}\) to \(70^{\circ}\mathrm{C}\) in 4.8 minutes. The room temperature is \(20^{\circ}\mathrm{C}\). Applying Newton's law of cooling, the time needed to cool it further by \(10^{\circ}\mathrm{C}\) should be nearly:

One kilogram of ice at \(0^\circ \mathrm{C}\) is mixed with one kilogram of water at \(80^\circ \mathrm{C}.\) The final temperature of the mixture will be: (Take: Specific heat of water = \(4200\) J kg^-1 K^-1, latent heat of ice\(=336\) kJ kg^-1)
1. \(0^\circ \mathrm{C}\)
2. \(50^\circ \mathrm{C}\)
3. \(40^\circ \mathrm{C}\)
4. \(60^\circ \mathrm{C}\)

The value of the coefficient of volume expansion of glycerin is \(5\times10^{-4}\) K^-1. The fractional change in the density of glycerin for a temperature increase of \(40^\circ \mathrm{C}\) will be:

Steam at \(100^{\circ}\mathrm{C}\) is injected into 20 g of \(10^{\circ}\mathrm{C}\) water. When water acquires a temperature of \(80^{\circ}\mathrm{C}\), the mass of water present will be: (Take specific heat of water =1 cal g^-1 \(^\circ\)C^-1 and latent heat of steam = 540 cal g^-1)
1. 24 g                       
2. 31.5g
3. 42.5 g                   
4. 22.5 g

The temperature of a body falls from \(50^{\circ}\mathrm{C}\)$\mathrm{}$ to \(40^{\circ}\mathrm{C}\)$\mathrm{}$ in 10 minutes. If the temperature of the surroundings is \(20^{\circ}\mathrm{C}\)$\mathrm{,\; t}$hen the temperature of the body after another 10 minutes will be:
1. \(36.6^{\circ}\mathrm{C}\)          
2. \(33.3^{\circ}\mathrm{C}\)$\mathrm{}$
3. \(35^{\circ}\mathrm{C}\)             
4. \(30^{\circ}\mathrm{C}\)$\mathrm{}$

Two rods (one semi-circular and the other straight) of the same material and of the same cross-sectional area are joined as shown in the figure. Points A and B are maintained at different temperatures. The ratio of the heat transferred through a cross-section of a semi-circular rod to the heat transferred through a cross-section of a straight rod at any given point in time will be:
     
1. 2: $\mathrm{\pi }$ 

2. 1: 2

3. $\mathrm{\pi }$: 2

4. 3: 2

The plots of intensity versus wavelength for three black bodies at temperatures ${\mathrm{T}}_1$, ${\mathrm{T}}_2$ and ${\mathrm{T}}_3$ respectively are as shown. Their temperatures are such that:
         

A substance is in solid form at \(0^{\circ}\mathrm{C}\). The amount of heat added to this substance and its temperature are plotted in the following graph. If the relative specific heat capacity of the solid substance is 0.5, from the graph, the specific latent heat of the melting process is: (Specific heat capacity of water = 1000 cal kg^-1 K^-1 )
      
1. 60000 cal kg^-1 
2. 40000 cal kg^-1
3. 10000 cal kg^-1 
4. 20000 cal kg^-1