JEE Advanced 2025 Sample Questions
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- A projectile is thrown from point O on the ground at an angle of 45° from the vertical and with a speed of 5 √2 m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, t seconds after the splitting, falls to the ground at a distance x meters from point O. The acceleration due to gravity g = 10 m/s2.
a. The value of t is ________. Answer: 0.5
H = u2sin2θ/2g
- The value of x is ________. Answer: 7.5
X = 3R/2 as Xcm = R
R = u2sin2θ/g
= 50/10 = 5
⇒X = 3R/2 = 15/2 = 7.5 m
2. An aqueous solution is prepared by dissolving 0.1 mol of an ionic salt in 1.8 kg of water at 35ºC. The salt remains 90% dissociated in the solution. The vapour pressure of the solution is 59.724 mm of Hg. The vapour pressure of water at 35ºC is 60.000 mm of Hg. The number of ions present per formula unit of the ionic salt is _______.
Answer (5)
Sol. Number of ions present per formula unit of ionic salt = x
Van ‘t Hoff factor (i) = 0.9 × x + 0.1 × 1 (Assuming 90% dissociation)
∴ Relative lowering in vapour pressure = Mole fraction of solute
3. The correct option(s) related to the extraction of iron from its ore in the blast furnace operating in the temperature range 900 – 1500 K is(are)
(A) Limestone is used to remove silicate impurities.
(B) Pig iron obtained from the blast furnace contains about 4% carbon.
(C) Coke (C) converts CO2 to CO.
(D) Exhaust gases consist of NO2 and CO.
Answer (A, B, C)
Sol. (A) Limestone is added to remove silica as an impurity.
(B) Pig iron obtained from a blast furnace contains 4% carbon and many other impurities (eg. S, P, Si, Mn) in small amounts.
(C) Coke (C) converts CO2 into CO. C + CO2 → 2CO
(D) Exhaust gases consist of CO and CO2.
Hence, (A, B, C) are correct.
4. Let l1, l2, …, l100 be consecutive terms of an arithmetic progression with common difference d1, and let w1, w2, …, w100 be consecutive terms of another arithmetic progression with common difference d2, where d1d2 = 10. For each i = 1, 2, …, 100, let Ri be a rectangle with length li, width wi and area Ai. If A51 − A50 = 1000, then the value of A100 − A90 is ____________.
Answer (18900)
Sol. For A.P. l1, l2, … l100
Let T1 = a and common difference = d1 and similarly for A.P. w1, w2, … w100
T1 = b and common difference = d2
A51 – A50 = l51w51 – l50w50
= (a + 50d1)(b + 50d2) – (a + 49d1)(b + 49d2)
= 50bd1 + 50ad2 +2500d1d2 – 49ad2 – 49bd1 – 2401d1d2
= bd1 + ad2 + 99d1d2 = 1000
∴ bd1 + ad2 = 10 …(i) (As d1d2 = 10)
∴ A100 – A90 = l100w100 – l90w90
= (a + 99d1)(b + 99d2) – (a + 89d1)(b + 89d2)
= 99bd1 + 99ad2 + 992d1d2 – 89bd1 – 89ad2 – 892d1d2
= 10(bd1 + ad2) + 1880d1d2
= 10(10) + 18800
= 18900
5. Let the functions f: (−1,1) → R and g: (−1, 1) → (−1, 1) be defined by f(x) = |2x − 1| + |2x + 1|and g (x) = x − [x], where [x] denotes the greatest integer less than or equal to x. Let fog: (−1,1) → R be the composite function defined by (fog)(x) = f(g(x)). Suppose C is the number of points in the interval (−1, 1) at which fog is NOT continuous, and suppose d is the number of points in the interval (−1,1) at which fog is NOT differentiable. Then the value of c + d is _____
Answer: 4
Solution:
f(x) = |2x − 1| + |2x + 1|
f(x) = -4x at x ≤ -1/2
f(x) = 2 at (-1/2) < x < (1/2)
f(x) = 4x at x ≥ ½
g(x) = x [x] = {x}
Now,
Now check
fog is not continuous at x = 0 only => c = 1
fog is not differentiable at x = (-1/2), 0, (1/2), => d = 3
So, c + d = 4
