A ball is released from rest and rolls down an inclined plane, as shown in the following figure, requiring 4 s to cover a distance of 100 cm along the plane : Which one of the following is the correct value of angle θ that the plane makes with the horizontal? (g = 1000 cm/s²)

Q: What is the answer to NDA (II) 2018 GAT Q111?

C. Correct — C, θ = sin⁻¹ (1/80). The printed figure shows a straight incline rising from a horizontal base, with the angle between the two marked θ and the ball at the upper end, so the 100 cm is measured along the sloping surface itself, not along the ground. The ball starts from rest, so u = 0 and the distance formula s = ut + ½at² reduces to s = ½at². With s = 100 cm and t = 4 s: 100 = ½ × a × 16, so a = 200/16 = **12·5 cm/s²**. For a body released on an incline the acceleration along the slope is the component of gravity in that direction, a = g sin θ. Taking g = 1000 cm/s² as the question specifies — the units are already in centimetres, which is why g is given in cm/s² — sin θ = a/g = 12·5 / 1000 = 1/80, so θ = sin⁻¹ (1/80).

(a)θ = sin⁻¹ (1/9·8)
(b)θ = sin⁻¹ (1/20)
(c)θ = sin⁻¹ (1/80)
(d)θ = sin⁻¹ (1/100)

Answer

Why

Correct — C, θ = sin⁻¹ (1/80). The printed figure shows a straight incline rising from a horizontal base, with the angle between the two marked θ and the ball at the upper end, so the 100 cm is measured along the sloping surface itself, not along the ground.

The ball starts from rest, so u = 0 and the distance formula s = ut + ½at² reduces to s = ½at². With s = 100 cm and t = 4 s:

100 = ½ × a × 16, so a = 200/16 = 12·5 cm/s².

For a body released on an incline the acceleration along the slope is the component of gravity in that direction, a = g sin θ. Taking g = 1000 cm/s² as the question specifies — the units are already in centimetres, which is why g is given in cm/s² —

sin θ = a/g = 12·5 / 1000 = 1/80, so θ = sin⁻¹ (1/80).

Why the others are wrong

(a)
θ = sin⁻¹ (1/9·8) — This mixes the unit systems. It divides the acceleration by 9·8 m/s² while the distance was worked in centimetres. The question deliberately states g = 1000 cm/s² so that every quantity stays in cm — using 9·8 here is a factor-of-100 error.
(b)
θ = sin⁻¹ (1/20) — This follows from a = 50 cm/s², which is what s = ½at² gives if t² is taken as 4 instead of 16 — squaring the time is the step being tested.
(d)
θ = sin⁻¹ (1/100) — This comes from setting a = 10 cm/s², effectively using s = at² or reading g as 10 000 cm/s². The correct acceleration from s = ½at² is 12·5 cm/s², giving 1/80 rather than 1/100.

Concept

A body released from rest on a smooth incline accelerates down the slope under the component of gravity along it, a = g sin θ. The component perpendicular to the slope is balanced by the normal reaction and does not drive the motion. Because the start is from rest, the kinematic relation s = ut + ½at² collapses to s = ½at², so acceleration follows directly from a distance and a time.

Two things decide this item. The first is squaring the time: t = 4 s gives t² = 16, and using 4 instead produces option (b). The second is the unit system — the question hands you g in cm/s² precisely because the distance is in centimetres, and reaching for 9·8 instead produces option (a). Work the acceleration first from the kinematics, then convert it to an angle through sin θ = a/g.

Key facts

From rest, s = ½at², so a = 2s/t².
Here a = 2(100)/4² = 12·5 cm/s².
On an incline the driving acceleration is a = g sin θ.
sin θ = a/g = 12·5/1000 = 1/80.
g is given as 1000 cm/s² so the whole problem stays in centimetres.

From distance and time to the angle

Released from rest→u = 0, so s = ½at²

Substitute s = 100 cm, t = 4 s→100 = ½ × a × 16

Acceleration along the slope→a = 200/16 = 12·5 cm/s²

On an incline, a = g sin θ→sin θ = 12·5 / 1000 = 1/80

Square the time and keep everything in centimetres — θ = sin⁻¹ (1/80), option (c).

Study next

Motion on an inclined plane and the g sin θ componentKinematic equations for motion from restResolving weight into components along and normal to a surfaceConsistency of units in mechanics problems

Common traps

Forgetting to square the time in s = ½at².
Switching to g = 9·8 m/s² when the question supplies g in cm/s².
Treating the 100 cm as a horizontal distance instead of a distance along the slope.
Dropping the factor of ½ and using s = at².

NDA gives a distance along an incline and the time taken from rest, then asks for the angle — get a from 2s/t², then read the angle off sin θ = a/g in the units the question supplies.

Related PYQs

UPSC_1997_GS1_Q1341997

A smooth inclined plane is inclined at an angle θ with the horizontal as shown in the figure. A body starts from rest and slides down the inclined surface. The time taken by the body to reach the bottom is

Answer(c) √(2l / (g sin θ))

The same smooth-incline setup in the UPSC Prelims, asking for the time to reach the bottom instead of the angle — both rest on a = g sin θ with the distance measured along the slope.

Practice

practice — not a real PYQ
A body starts from rest and covers 50 cm along a smooth incline in 2 s. What is its acceleration along the slope?
- (a)12·5 cm/s²
- (b)25 cm/s²
- (c)50 cm/s²
- (d)100 cm/s²
Answer(b) 25 cm/s² — a = 2s/t² = 2(50)/4 = 25 cm/s².
practice — not a real PYQ
A block slides down a smooth plane inclined at 30° to the horizontal. Its acceleration along the plane is
- (a)g
- (b)g/2
- (c)g√3/2
- (d)zero
Answer(b) g/2 — a = g sin 30° = g × ½.