Home Maths JEE - Advanced Previous Year Paper JEE-Advanced-2017-Paper-2 SECTION 2 [Maximum Marks: 28] • This section…
Maths JEE - Advanced Previous Year Paper JEE-Advanced-2017-Paper-2 Single Correct MCQ
Published on: August 12, 2026

SECTION 2 [Maximum Marks: 28]

• This section contains SEVEN questions

• Each question has FOUR options

A
, , and and Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system
B
, will result in -2 marks, as a wrong option is also darkened. The correct statement(s) about surface properties is(are) The critical temperatures of ethane and nitrogen are 563 K and 126 K, respectively. The adsorption of ethane will be more than that of nitrogen on same amount of activated charcoal at a given temperature
C
and and Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion medium
D
. ONE OR MORE THAN ONE of these four option(s) is(are) correct. • For each question, marks will be awarded in one of the following categories: Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened. Zero Marks : 0 If none of the bubbles is darkened Negative Marks : -2 In all other cases • For example, if are all the correct options for a question, darkening all these three will get +4 marks; darkening only will result in +2 marks; and darkening Brownian motion of colloidal particles does not depend on the size of the particles but depends on viscosity of the solution

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Text Solution

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The correct answer is:
CHECK THE SOLUTION.

(a, b)

In adsorption process both AH & AS is - ve. Higher the critical temperature of a gas higher the extent of adsorption.

Cloud is not an emulsion.

Brownian motion depends on the size of the particles.

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