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)
θ = sin⁻¹ (1/9•8)
θ = sin⁻¹ (1/20)
θ = sin⁻¹ (1/80)
θ = sin⁻¹ (1/100)
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC NDA-2 – 2018
– The distance covered along the plane is s = 100 cm in time t = 4 s.
– The motion along the inclined plane is uniformly accelerated. The formula for distance is s = ut + (1/2)at².
– Substitute the values: 100 cm = (0)(4 s) + (1/2)a(4 s)².
– 100 = (1/2)a(16) => 100 = 8a.
– The acceleration down the inclined plane is a = 100 / 8 = 12.5 cm/s².
– The acceleration of an object rolling down an inclined plane (without slipping, but here it’s just about acceleration along the plane component of gravity) is a = g sin θ, where g is the acceleration due to gravity and θ is the angle of inclination.
– Given g = 1000 cm/s².
– So, 12.5 = 1000 sin θ.
– sin θ = 12.5 / 1000 = 125 / 10000 = 1 / 80.
– θ = sin⁻¹ (1/80).
– The value of g is given as 1000 cm/s², which is equivalent to 10 m/s².