Ncert Exercises & Solutions

A body of mass $10 ~\text{Kg}$ is acted upon by two perpendicular forces, $6 ~\text{N}$ and $8 ~\text{N}$ . The resultant acceleration of the body is:

(a)	$1~\text{ms}^{-2}$ at an angle of $\text {tan}^{-1} \left(\dfrac{4}{3}\right )$ w.r.t. $6 ~\text{N}$ force.
(b)	$0.2~\text{ms}^{-2}$ at an angle of $\text {tan}^{-1} \left(\dfrac{3}{4}\right )$ w.r.t. $8 ~\text{N}$ force.
(c)	$1~\text{ms}^{-2}$ at an angle of $\text {tan}^{-1} \left(\dfrac{3}{4}\right )$ w.r.t. $8 ~\text{N}$ force.
(d)	$0.2~\text{ms}^{-2}$ at an angle of $\text {tan}^{-1} \left(\dfrac{3}{4}\right )$ w.r.t. $6 ~\text{N}$ force.

Choose the correct option:
1. (a), (c)
2. (b), (c)
3. (c), (d)
4. (a), (b), (c)

(1) Hint: The resultant acceleration will be in the direction of the resultant force.

Step 1: Find the resultant force and the magnitude of the acceleration.

Consider the adjacent diagram
Given, mass = m = 10 kg

\begin{matrix} Resultant force = F = \sqrt{F_{1}^{2} + F_{2}^{2}} = \sqrt{36 + 64} = 10 N a = \frac{F}{m} = \frac{10}{10} = 1 m / s^{2}; along R \end{matrix}

$\begin{array}{l} Resultant force = F = \sqrt{F_{1}^{2} + F_{2}^{2}} = \sqrt{36 + 64} \\ = 10 N \\ a = \frac{F}{m} = \frac{10}{10} = 1 m / s^{2}; along R \end{array}$

Step 2: Find the direction of the acceleration.

\begin{matrix} Let θ_{1} be angle between R and F_{1} \begin{matrix} tan θ_{1} = \frac{8}{6} = \frac{4}{3} θ_{1} = {tan}^{- 1} (4 / 3) w . r . t . F_{1} = 6 N \end{matrix} \end{matrix}

$\begin{array}{l} Let θ_{1} be angle between R and F_{1} \\ \begin{matrix} \tan θ_{1} = \frac{8}{6} = \frac{4}{3} \\ θ_{1} = \tan^{- 1} (4 / 3) w . r . t . F_{1} = 6 N \end{matrix} \end{array}$

Let θ_{2} be angle between F and F_{2}

$Let θ_{2} be angle between F and F_{2}$

tan θ_{2} = \frac{6}{8} = \frac{3}{4}

$\tan θ_{2} = \frac{6}{8} = \frac{3}{4}$

θ_{2} = {tan}^{- 1} (\frac{3}{4}) w . r . t . F_{2} = 8 N

$θ_{2} = \tan^{- 1} (\frac{3}{4}) w . r . t . F_{2} = 8 N$

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