1

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

A metal ball of mass 0.1 kg is heated upto 500oC and dropped into a vessel of heat capacity 800 JK–1 and containing 0.5 kg water. The initial temperature of water and vessel is 30oC. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, 4200 Jkg–1 and 400 Jkg–1 K–1
A
20%
B
25%
C
15%
D
30%

## Explanation

0.1 $\times$ 400 $\times$ (500 $-$ T) = 0.5 $\times$ 4200 $\times$ (T $-$ 30) + 800 (T $-$ 30)

$\Rightarrow$  40(500 $-$ T) = (T $-$ 30) (2100 + 800)

$\Rightarrow$  20000 $-$ 40T = 2900 T $-$ 30 $\times$ 2900

$\Rightarrow$  20000 + 30 $\times$ 2900 = T(2940)

T = 30.4oC

${{\Delta T} \over T} \times 100$ = ${{6.4} \over {30}} \times 100$

$\simeq$ 20%
2

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

A thermometer graduated according to a linear scale reads a value x0 when in contact with boiling water, and x0/3 when in contact with ice. What is the temperature of an object in oC, if this thermometer in the contact with the object reads x0/2 ?
A
60
B
35
C
25
D
40

## Explanation $\Rightarrow$   ToC = ${{{x_0}} \over 6}$ & $\left( {{x_0} - {{{x_0}} \over 3}} \right)$ = (100 $-$ 0oC)

x0 = ${{300} \over 2}$

$\Rightarrow$  ToC = ${{150} \over 6}$ = 25oC
3

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

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is increased by $\Delta$T. The amount of heat absorbed by gas is (R is gas constant) :
A
${1 \over 2}$ KR$\Delta$T
B
${1 \over 2}$ R$\Delta$T
C
${3 \over 2}$ R$\Delta$T
D
${2K \over 3}$ $\Delta$T

## Explanation

VT = K

$\Rightarrow$  V$\left( {{{PV} \over {nR}}} \right)$ = k $\Rightarrow$ PV2 = K

$\because$  C = ${R \over {1 - x}} +$ Cv (For polytropic process)

C = ${R \over {1 - 2}} + {{3R} \over 2}$ = ${R \over 2}$

$\therefore$  $\Delta$Q = nC $\Delta$T

= ${R \over 2} \times \Delta$T
4

### JEE Main 2019 (Online) 12th January Morning Slot

An ideal gas occupies a volume of 2m3 at a pressure of 3 $\times$ 106 Pa. The energy of the gas is :
A
6 $\times$ 104 J
B
9$\times$ 106 J
C
3 $\times$ 102 J
D
108 J

## Explanation

Energy = ${1 \over 2}$ nRT = ${f \over 2}$PV

= ${f \over 2}$ (3 $\times$ 106) (2)

= f $\times$ 3 $\times$ 106

Considering gas is monoatomic i.e. f = 3

E. = 9 $\times$ 106 J