JEE Advanced Mix Test 59

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JEE Advanced Mix Test 59

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Question 1
4 / -1

A.DC circuit consisting of two cells of emf V and 2V having no internal resistance are connected with two capacitors of capacity C and 2C and four resistors R, R, 2R and 2R as shown in figure. The ammeter and voltmeter used in the circuit are ideal.

The reading of the ammeter as soon as the switch is closed is

A
v/R

B
v/2R

C
3v/R

D
Zero

Solution

Capacitors will offer no resistance and hence the circuit becomes

Clearly reading of ammeter will be zero.
Question 2
4 / -1

An ideal diatomic gas is expanded so that the amount of heat transferred to the gas is equal to the decrease in its internal energy.

The molar specific heat of the gas in this process is given by C whose value is

A

B

C
2R

D
5R

Solution

It is given that there is decrease of the internal energy while the gas expands absorbing some heat, which numerically equal to the decrease in internal energy.

Hence, dQ = −dU

Thus, dQ = dW + dU = −dU

⇒ dW = −2dU (remembering that both dW and dU are negative)

⇒ PdV = 2CdT [dU = −CdT] ...... (1)

where C is the molar specific heat of the gas

Since the gas is ideal, PV = RT ....... (2)

From (2) P = RT/v
Question 3
4 / -1

A reversible cyclic process involves 6 steps. In step-1 and 3, the system absorbs 500 J and 800 J of heat from a heat reservoir at temperature 250 K (T₁) and 200 K (T₃) respectively. Step-2, 4 and 6 are adiabatic such that the temperature of one reservoir changes to that of next.

Total work done by the system in whole process is 700 J.

If total entropy change during steps-1,3 and 5 be ’x’ J/K, then the value of is .....

A
0.80

B
0.100

C
0.50

D
0.10

Solution
Question 4
4 / -1

A reversible cyclic process involves 6 steps. In step-1 and 3, the system absorbs 500 J and 800 J of heat from a heat reservoir at temperature 250 K (T₁) and 200 K (T₃) respectively. Step-2, 4 and 6 are adiabatic such that the temperature of one reserv oir changes to that of next.

Total work done by the system in whole process is 700 J.

If during step-5, the system exchanges heat from a reserv oir at temperature ‘T₅ K’, then value of T₁/T₂ is:

A
2.50

B
4.50

C
1.10

D
5.50

Solution
Question 5
4 / -1

Two beaker are placed in a sealed flask. Beaker A initially contained 0.15 mol of napthalene (non-volatile) in 117 g of benzene and beaker B initially contained 31 g of an unknown compound (non-volatile, non-electrolytic) in 117 g of benzene. At equilibrium, beaker A is found to have lost 7.8 g of weight. Assume ideal behaviour of both solutions to answer the following question.

The molar mass of solute in solution B is closest to:

A
60

B
30

C
181

D
342

Solution
Question 6
4 / -1

Two beaker are placed in a sealed flask. Beaker A initially contained 0.15 mol of napthalene (non-volatile) in 117 g of benzene and beaker B initially contained 31 g of an unknown compound (non-volatile, non-electrolytic) in 117 g of benzene. At equilibrium, beaker A is found to have lost 7.8 g of weight. Assume ideal behaviour of both solutions to answer the following question.

Now beaker B, at equilibrium, is replaced by another beaker C which contain 31 g of same solute as in B, but in different solvent X. Also moles of X in solution C is same as mol of benzene in solution B at equilibrium. When the container containing beakers A and C is closed again and allowed to attain equilibrium, beaker C has lost some weight. Which of the following can be said with guarantee ?

A
The weight loss in C will be 7.8 g

B
Vapour pressure of X is greater than the vapour pressure of benzene

C
Moles of benzene migrated from C to A will be equal to the moles of benzene migrated from A to B in the first experiment

D
Boiling point of solution C is greater than boiling point of solution A

Solution

Relative lowering in vapour pressure is less over beaker C i.e., vapour pressure of X is more than vapour pressure of benzene.
Question 7
4 / -1

a, b, c are distinct and |a| + |b| = |c|, also |a + bω + cω²| is minimum (where ω is the complex cube root of unity). Now consider the lines having direction cosines connected by al + bm + cn = 0 and l² + m² - n² = 0 If a·b·c = 0, then angle between the lines is q, then 1/2 tan²θ is

A
1.50

B
2.50

C
3.50

D
4.50

Solution

Given: |a| + |b| = |c|, a + bω + cω² minimized, ω³ = 1, and a · b · c = 0.

Lines satisfy: al + bm + cn = 0, l² + m² + n² = 1.

If c = 0, then |a| = |b|, and al + bm = 0. Minimize |a + bω|: set b = –aω, but |a| = |–aω| = |a|, so |c| = 0, contradicting distinctness unless a = –b.

Assume a = 0, then |b| = |c|, and bm + cn = 0. Lines: solve for direction cosines.

Solving gives two lines with angle θ = 60°, so:

tan θ = tan 60° = √3,

tan² θ = 3,

(1/2) tan² θ = (3/2) = 1.50.
Question 8
4 / -1

a, b, c are distinct and |a| + |b| = |c|, also |a + bω + cω²| is minimum (where ω is the complex cube root of unity) . Now consider the lines having direction cosines connected by al + bm + cn = 0 and l² + m² - n² = 0 Total number of ordered triplets (a, b, c) are

A
5.00

B
9.00

C
10.00

D
8.00

Solution

Conditions: |a| + |b| = |c|, minimize |a + bω + cω²|, a, b, c distinct.

Let |a| = x, |b| = y, |c| = x + y. Minimize:

|a + bω + cω²|² = (a + bω + cω²)(a + bω² + cω).

Optimal when a + bω + cω² = 0, but check real solutions.

Possible signs: a = ±x, b = ±y, c = ±(x + y), distinct values.

Ordered triplets: Consider a, b, c > 0, then negate combinations ensuring distinctness.

Total: 8 triplets (e.g., permutations of signs and choices like a = 0, b = c).

Triplets satisfying |a| + |b| = |c|, a + bω + cω² minimized, a ≠ b ≠ c,

Total = 8.
Question 9
4 / -1

Let the line at two points A and B. If P be any v ariable point on the ellipse The number of points P on ellipse is three, when the area of the is equal to

A
1.25

B
2.25

C
1.50

D
1.80

Solution
Question 10
4 / -1

Let the line at two points A and B. If P be any v ariable point on the ellipse The number of point P on ellipse is one, when the area of is equal to

A
1.60

B
1.80

C
1.50

D
2.50

Solution