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Question: A car of mass 2 metric tons is moving up a smooth inclined plane with a velocity of 42kmph. If the p...

A car of mass 2 metric tons is moving up a smooth inclined plane with a velocity of 42kmph. If the power of the engine is 8 KW, the angle of inclination of the inclined plane is

& \text{A}\text{. ta}{{\text{n}}^{-1}}\left( \dfrac{1}{49} \right) \\\ & \text{B}\text{. co}{{\text{s}}^{-1}}\left( \dfrac{48}{49} \right) \\\ & \text{C}\text{. co}{{\text{s}}^{-1}}\left( \dfrac{1}{49} \right) \\\ & \text{D}\text{. si}{{\text{n}}^{-1}}\left( \dfrac{1}{49} \right) \\\ \end{aligned}$$
Explanation

Solution

At first take out all the values that we can get from the question, then solve to find how we can find the force of the car when it's in an inclined plane. Then from the formula for power of the car we will get the required result.

Formula used:
F=mgsinθF=mg\sin \theta
P=F×vP=F\times v

Complete answer:
According to the question, the mass of the car is 2 metric tons, or 2000Kg.
The car is moving up a smooth inclined plane with a velocity of 42 km/h.
Power of the engine is 4 KW. So this is 4000W
As the car is moving on a plane then, the plane must be at an angle so,
Let us assume the angle of inclination asθ\theta .
Therefore the force with which the car must move is,
F=mgsinθF=mg\sin \theta
F=2000×10sinθF=2000\times 10\sin \theta
F=20000sinθF=20000\sin \theta
Therefore according to the concept we know that power is P=F×vP=F\times v
8000=20000sinθ×4200036008000=20000\sin \theta \times \dfrac{42000}{3600} ,
So on solving this we get,
sinθ=(6175)\sin \theta =\left( \dfrac{6}{175} \right),
θ=sin1(6175)\theta ={{\sin }^{-1}}\left( \dfrac{6}{175} \right).

Therefore the correct option should be,
Option D, as that is the closest to the answer we got.

Note:
In the formula F=mgsinθF=mg\sin \theta , F is the force, m is the mass of the car, g is the acceleration due to gravity and θ\theta is the angle at which the inclined plane is angled. In the formula of power P=F×vP=F\times v, F is the force and v is the velocity. In this step 8000=20000sinθ×4200036008000=20000\sin \theta \times \dfrac{42000}{3600},420003600\dfrac{42000}{3600} is actually the velocity of the car as it was given in the question in KMPH form so 42KM is changed to 42000m and 1hr is 3600second.