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1

JEE Main 2019 (Online) 12th January Morning Slot

Given
Gas H2 CH4 CO2 SO2
Critical
Temperature / K
33 190 304 630

On the basis of data given above, predict which of the following gases shows least adsorption on a definite amount of charcoal ?
A
SO2
B
CH4
C
H2
D
CO2

Explanation

Adsorption $$\propto$$ Tc

Therefore, H2 gas shows least adsorption on a definite amount of charcoal.
2

JEE Main 2019 (Online) 11th January Evening Slot

Among the colloids cheese (C), milk (M), and smoke (S), the correct combination of the dispersed phase and dispersion medium, respectively is :
A
C : liquid in solid ; M : liquid in liquid ; S : solid in gas
B
C : liquid in solid ; M : liquid in solid ; S : solid in gas
C
C : solid in liquid ; M : liquid in liquid ; S : gas in solid
D
C : solid in liqid ; M : solid in liquid ; S : solid in gas

Explanation

Cheese (C) is liquid dispersed in solid.

Milk (M) is liquid dispersed in liquid.

Smoke (S) is solid dispersed in gas.
3

JEE Main 2019 (Online) 11th January Evening Slot

The radius of the largest sphere which fits properly at the centre of the edge of a body centred cubic unit cell is : (Edge length is represented by 'a')
A
0.134 a
B
0.067 a
C
0.047 a
D
0.027 a

Explanation

a = 2(R + r)

$$\Rightarrow$$ $${a \over 2} = \left( {R + r} \right)$$ ......(1)

For bcc, $$\sqrt 3 a$$ = 4R

Using (i) and (ii)

$${a \over 2} = \left( {{{a\sqrt 3 } \over 4} + r} \right)$$

$$\Rightarrow$$ r = $$a\left( {{{2 - \sqrt 3 } \over 4}} \right)$$

$$\Rightarrow$$ r = 0.067 a
4

JEE Main 2019 (Online) 11th January Morning Slot

A solid having density of 9$$\times$$103 kg m–3 forms face centred cubic crystals of edge length $$200\sqrt 2$$ pm. What is the molar mass of the solid?
[Avogadro constant $$\cong$$ 6 $$\times$$ 1023 mol–1 , $$\pi$$ $$\cong$$ 3]
A
0.0305 kg mol–1
B
0.4320 kg mol–1
C
0.0216 kg mol–1
D
0.0432 kg mol–1

Explanation

$$\rho$$ = $${{Z \times M} \over {{a^3} \times {N_A}}}$$

$$\Rightarrow$$ 9$$\times$$103 = $${{4 \times M} \over {\left( {200\sqrt 2 \times {{10}^{ - 12}}} \right) \times 6 \times {{10}^{23}}}}$$

$$\Rightarrow$$ M = 0.0305 kg mol–1

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