Question

The resultant of two forces $2P$ and $\sqrt 2 P$ is $\sqrt {10} P$. The angle between the forces is
(A) ${30^ \circ }$
(B) ${60^ \circ }$
(C) ${45^ \circ }$
(D) ${90^ \circ }$

Answer 1

Hint
In the above given problem the angle between the two forces which are of vector quantity can be found by using the formula derived to find the equation for the resultant between two vectors of the respective forces.
The formula for the angle between the forces is given by;
$\Rightarrow R = \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta } $
Where, $R$ denotes the resultant between two vectors, $P$ denotes the magnitude of the first vector, $Q$ denotes the magnitude of the second vector, $\theta $ denotes the angle between the two forces.

Complete step by step answer
The data given in the problem are:
The magnitude of the first vector is $P = 2P$
The magnitude of the second vector is $Q = \sqrt 2 P$
the resultant between two vectors is, $R = \sqrt {10} P$
The formula for the angle between the forces is given by;
$\Rightarrow R = \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta } $
Substitute the values of the magnitude of the first vector, the values of the magnitude of the second vector is and the resultant between two vectors in the above equation;
$\Rightarrow \sqrt {10} P = \sqrt {{{\left( {2P} \right)}^2} + {{\left( {\sqrt 2 P} \right)}^2} + 2 \times \left( {2P} \right) \times \left( {\sqrt 2 P} \right) \times \cos \theta } $
$\Rightarrow \sqrt {10} P = \sqrt {4{P^2} + 2{P^2} + 4\sqrt 2 {P^2}\cos \theta } $
By squaring on both sides, we get;
$\Rightarrow 10{P^2} = 4{P^2} + 2{P^2} + 4\sqrt 2 {P^2}\cos \theta $
$\Rightarrow 10{P^2} - 4{P^2} - 2{P^2} = 4\sqrt 2 {P^2}\cos \theta $
$\Rightarrow 4{P^2} = 4\sqrt 2 {P^2}\cos \theta $
Cancelling the like terms and equating the above equation;
$\Rightarrow 4{P^2} = 4\sqrt 2 {P^2}\cos \theta $
$\Rightarrow \cos \theta = \dfrac{4}{{4\sqrt 2 }} $
$\Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right) $
$\Rightarrow \theta = {45^ \circ } $
Therefore, the angle between the two forces is found as $\theta = {45^ \circ }$.
Hence the option (C) $\theta = {45^ \circ }$ is the correct answer.

Note
The resultant value is the vector of the sum of two or more vectors. It is the product of adding two or more vectors together. When there are displacement vectors which are added, the product is a resultant displacement. But any two vectors can be added as long as they are the same vector quantity.