1

### JEE Main 2019 (Online) 11th January Morning Slot

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TVx = constant, then x is :
A
${5 \over 3}$
B
${2 \over 5}$
C
${3 \over 5}$
D
${2 \over 3}$

## Explanation

For adiabatic process : TV$\gamma $$-1 = constant For diatomic process : \gamma$$-$1 = ${7 \over 5} - 1$

$\therefore$  x = ${2 \over 5}$
2

### JEE Main 2019 (Online) 11th January Morning Slot

Ice at –20oC is added to 50 g of water at 40oC. When the temperature of the mixture reaches 0oC, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water = 4.2J/g/oC Specific heat of Ice = 2.1J/g/oC Heat of fusion of water at 0oC= 334J/g)
A
100 g
B
60 g
C
50 g
D
40 g

## Explanation

Let amount of ice is m gm.

According to principal of calorimeter heat taken by ice = heat given by water

$\therefore$  20 $\times$ 2.1 $\times$ m + (m $-$ 20) $\times$ 334

= 50 $\times$ 4.2 $\times$ 40

376 m = 8400 + 6680

m = 40.1
3

### JEE Main 2019 (Online) 11th January Evening Slot

Two rods A and B of identical dimensions are at temperature 30°C. If A is heated upto 180oC and B upto ToC, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then the value of T is
A
200oC
B
270oC
C
230oC
D
250oC

## Explanation

$\Delta {\ell _1} = \Delta {\ell _2}$

$\ell {\alpha _1}\Delta {T_1} = \ell {\alpha _2}\Delta {T_2}$

${{{\alpha _1}} \over {{\alpha _2}}} = {{\Delta {T_1}} \over {\Delta {T_2}}}$

${4 \over 3} = {{T - 30} \over {180 - 30}}$

$T = {230^o}C$
4

### JEE Main 2019 (Online) 11th January Evening Slot

When 100 g of a liquid A at 100oC is added to 50 g of a liquid B at temperature 75oC, the temperature of the mixture becomes 90oC. The temperature of the mixture, if 100 g of liquid A at 100oC is added to 50 g of liquid B at 50oC, will be :
A
60oC
B
70oC
C
85oC
D
80oC

## Explanation

100 $\times$ SA $\times$ [100 $-$ 90] = 50 $\times$ SB $\times$ (90 $-$ 75)

2SA = 1.5 SB

SA = ${3 \over 4}$SB

Now, 100 $\times$ SA $\times$ [100 $-$ T] = 50 $\times$ SB (T $-$ 50)

2 $\times$ $\left( {{3 \over 4}} \right)$ (100 $-$ T) = (T $-$ 50)

300 $-$ 3T = 2T $-$ 100

400 = 5T

T = 80

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