Laws of Motion

From the free body diagram,

$ R+F \sin 45^{\circ}=m g $

But, friction force, $f_s=\mu R$

$ \begin{aligned} \Rightarrow & R=\frac{f_s}{\mu} \Rightarrow R=\frac{F \cos 45^{\circ}}{\mu} \\ \Rightarrow & R=\frac{F}{\sqrt{2} \mu} \\ \therefore & \frac{F}{\sqrt{2} \mu}+\frac{F}{\sqrt{2}}=m g \\ & \frac{F}{\sqrt{2}}\left(\frac{1}{\mu}+1\right)=m g \\ \Rightarrow & F\left(\frac{\mu+1}{\mu}\right)=\sqrt{2} m g \\ & F\left(\frac{0.25+1}{0.25}\right)=\sqrt{2} \times \sqrt{2} \times 10 \\ & F \times 5=20 \\ \Rightarrow & F=4 \mathrm{~N} \end{aligned} $

$ \begin{aligned} F & =10 \mathrm{~N} \\ m & =1.5 \mathrm{~kg} \\ \mu & =0.2 \\ g & =10 \mathrm{~ms}^{-2} \\ t & =6 \mathrm{~s} \end{aligned} $

Frictional force $=\mu m g$

$ =0.2 \times 1.5 \times 10=3 \mathrm{~N} $

Net force $=$ Applied force

• -Frictional force

$ F_N=10-3=7 \mathrm{~N} $

Now,

$ \begin{aligned} F_N & =m a \Rightarrow 7=1.5 \times a \\ a & =\frac{7}{1.5}=4.67 \mathrm{~ms}^{-2} \end{aligned} $

Now, displacement in 6 seconds is given by

$ \begin{aligned} & s=u t+\frac{1}{2} a t^2 \\ & s=(0 \times 6)+\left[\frac{1}{2} \times 4.67 \times 6 \times 6\right] \\ & s=84.06 \mathrm{~m} \approx 84 \mathrm{~m} \end{aligned} $

∴ Work done $=$ Force × Displacement

$ =10 \times 84=840 \mathrm{~J} $

Total mass of truck and block

$ \begin{aligned} M^{\prime}=8+2 & =10 \mathrm{ton}=10 \times 1000 \mathrm{~kg} \\ & =10^4 \mathrm{~kg} \end{aligned} $

Acceleration produced due to application of 25 kN force,

$ a=\frac{F}{M^{\prime}}=\frac{25 \times 10^3}{10^4}=2.5 \mathrm{~m} / \mathrm{s}^2 $

Maximum frictional force acting on the block

$ \begin{aligned} F_{s m} & =\mu_s m g \\ & =0.3 \times 2000 \times 10=6000 \mathrm{~N} \end{aligned} $

The force required to decelerate the block with the same acceleration as the truck,

$ F^{\prime}=M_{\text {block }} \times a=2000 \times 2.5=5000 \mathrm{~N} $

Since, required force is less than the maximum static frictional force, thus, the block will not slide relative to the truck and the frictional force acting on the block will be equal to the required force to decelerate it with the truck's acceleration.

Thus, frictional force acting on the block $=5000 \mathrm{~N}$

Let $m_P$, $m_Q$, and $m_R$ be the masses of the bodies $P$, $Q$, and $R$.

Step 1: Write the formula for mass using force and acceleration.

We know that $F = m \times a$. Rearranging, $m = \frac{F}{a}$.

Step 2: Find the mass of each body.

For $P$, $m_P = \frac{F}{18}$

For $Q$, $m_Q = \frac{F}{9}$

For $R$, $m_R = \frac{F}{6}$

Step 3: Add all the masses to get total mass.

$m_{PQR} = m_P + m_Q + m_R$

So, $m_{PQR} = \frac{F}{18} + \frac{F}{9} + \frac{F}{6}$

Find a common denominator: $m_{PQR} = \frac{F}{18} + \frac{2F}{18} + \frac{3F}{18} = \frac{6F}{18} = \frac{F}{3}$

Step 4: Find the acceleration of the combined mass with the same force.

Acceleration, $a_{PQR} = \frac{F}{m_{PQR}} = \frac{F}{(F/3)}$

The $F$ cancels out: $a_{PQR} = 3\,\mathrm{m/s}^2$

Tension in the string connected to the 4 kg block $=T_2$

Tension in each side of the movable pulley (connected to 3 kg blocks) $=T_1$

Since, the pulley is ideal and frictionless, the net upward force on the pulley is

$ T_2=2 T_1 $

$4 g-T_2=4 a($ downward acceleration $a)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} & T_2=4 g-4 a \\ & T_1=3 g-3 a(\text { acceleration upwards } a) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

$ T_2=2 T_1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

From Eqs. (i) and (ii)

$ \begin{aligned} & 4 g-4 a=2(3 g-3 a) \\ & 4 g-4 a=6 g-6 a \\ & 2 a=2 g \\ \Rightarrow \quad & a=g \end{aligned} $

Substitute $a=g$ in Eq. (ii)

$ T_1=3 g-3 g=0(\text { invalid }) $

Assuming equilibrium $a=0$

Then, $T_1=3 g$

(From Eq. (ii))

$ \begin{aligned} & T_2=2 T_1 \\ & \frac{T_1}{T_2}=\frac{1}{2} \end{aligned} $

(From Eq. (iii))

When a man standing in a lift moving up with a retardation of $a \mathrm{~m} / \mathrm{s}^2$, then apparent weight $=m(g+a)$

$ \begin{aligned} & =60(9.8-2.8) \therefore a=-2.8 \mathrm{~m} / \mathrm{s}^2 \\ & =60 \times 7=420 \mathrm{~N} \end{aligned} $

Given:

Area of cross-section of the pipe, $ A = 5 \, \text{cm}^2 = 5 \times 10^{-4} \, \text{m}^2 $

Velocity of water, $ v = 5 \, \text{m/s} $

Density of water, $ \rho = 1 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3 $

To find the mass flow rate $\left(\frac{dm}{dt}\right)$, we use the formula:

$ \frac{dm}{dt} = \rho \cdot A \cdot v $

Calculating, we get:

$ \frac{dm}{dt} = 1000 \times 5 \times 10^{-4} \times 5 = 2.5 \, \text{kg/s} $

The water hits the block and rebounds with the same velocity but in the opposite direction. The change in velocity $(\Delta v)$ is:

$ \Delta v = v_{\text{final}} - v_{\text{initial}} = -5 - 5 = -10 \, \text{m/s} $

The force exerted by the water stream $F$ is calculated using the rate of change of momentum:

$ F = \frac{dm}{dt} \cdot \Delta v = 2.5 \times (-10) = -25 \, \text{N} $

(Note: The negative sign indicates the direction; for magnitude, we use the positive value.)

Thus, the force is $F = 25 \, \text{N}$.

Using Newton's Second Law:

$ F = ma \quad \Rightarrow \quad a = \frac{F}{m} = \frac{25}{5} = 5 \, \text{m/s}^2 $

Therefore, the initial acceleration of the block is $5 \, \text{m/s}^2$.

From net force acting on the body, From figure (i) and figure (ii)

$ \begin{array}{l} 20-f=m a \\\\ 20-f=m \times 6 \\\\ 30-f=m \times 11 \end{array} $

where $ f $ is frictional force (kinetic) From Eqs. (i) and (ii)

$ \begin{aligned} m & =2 \mathrm{~kg} \text { and } f=8 \mathrm{~N} \\\\ f & =\mu_{k} m g \\\\ \mu_{k} & =\frac{f}{m g}=\frac{8}{2 \times 10} \\\\ \mu_{k} & =0.4 \end{aligned} $

$ \mu_{k} $ coefficient of kinetic energy.

Given,

Mass of man $=50 \mathrm{~kg}$

Acceleration due to gravity

$ g=10 \mathrm{~m} / \mathrm{s}^2 $

Acceleration of lift $($ downward $)=g$

The net acceleration in case of lift, when lift is going downward

$ a_{\mathrm{net}}=g-a $

where $a$ is acceleration of lift.

$ \begin{aligned} & a_{\mathrm{net}}=g-g \quad(a=g \text { given }) \\ & a_{\mathrm{net}}=0 \end{aligned} $

So, the apparent weight is

$ \begin{aligned} & w=m a_{\text {net }} \\ & w=m \times 0 \Rightarrow w=0 \end{aligned} $

Given,

Weight of man when lift is not moving

$ =75 \mathrm{~kg} $

Lift is going downward with acceleration

$ =g=10 \mathrm{~m} / \mathrm{s}^2 $

The net acceleration for lift system going down

$ a_{\mathrm{net}}=g-a $

where $a$ is acceleration of lift.

$ \begin{gathered} a_{\mathrm{net}}=g-g \quad[a=g] \\ a_{\mathrm{net}}=0 \end{gathered} $

So, the weight inside it, when moving downward

$ \begin{aligned} & w=m a_{\mathrm{net}} \\ & w=0 \end{aligned} $

Thus, weight is zero.

Considering $w_2>w_1$, FBD is given below

The system is moved with acceleration $g$ upwards.

For motion of mass $m_2$, we have

$ w_2+m_2 g-T=m_2 a $

$ \begin{aligned} \Rightarrow & w_2+w_2-T =m_2 a \\ \Rightarrow & 2 w_2-T =m_2 a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

For motion of mass $m_1$, we have

$ \begin{array}{ll} & T-m_1 g-w_1=m_1 a \\ \Rightarrow & T-w_1-w_1=m_1 a \\ \Rightarrow & T-2 w_1=m_1 a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

From Eqs. (i), (ii), we get

$ \begin{aligned} & 2 w_2-T+T-2 w_1=m_2 a+m_1 a \\ & \Rightarrow \quad 2 w_2-2 w_1=a\left(m_2+m_1\right) \end{aligned} $

Since, $w_1=m_1 g$ and $w_2=m_2 g$

$ \begin{aligned} & \Rightarrow \quad 2 w_2-2 w_1=\left(\frac{w_2}{g}+\frac{w_1}{g}\right) \times a \\ & \Rightarrow \quad a=g\left(\frac{2 w_2-2 w_1}{w_1+w_2}\right) \end{aligned} $

Putting the value of a in Eq. (ii), we get

$ \begin{array}{rlrl} T-2 w_1 & =m_1 a \\ \Rightarrow & T-2 w_1 =m_1 g\left(\frac{2 w_2-2 w_1}{w_1+w_2}\right) \\ \Rightarrow & T =m_1 g\left(\frac{2 w_2-2 w_1}{w_1+w_2}\right)+2 w_1 \\ \Rightarrow & T =\left(\frac{4 w_1 w_2}{w_1+w_2}\right) g \quad\left(\because w_1=M_1 g\right) \end{array} $

Given, weight of body, $w=50 \mathrm{~N}$

Force $=28.28 \mathrm{~N}$

frictional force, $f=28.25 \times \cos 45^{\circ}$

$ =28.25 \times \frac{1}{\sqrt{2}}=20 \mathrm{~N} $

Normal reaction,

$ \begin{array}{r} R=50-28.2 \sin 45^{\circ} \\ =50-20=30 \mathrm{~N} \end{array} $

Hence, option (b) is correct.

Given,

Initially the person is at rest, after that start walking toward east.

Person does not skid or slip.

We know that, when we walk, our feet are pushing backward and therefore friction opposes this motion of pushing.

So, the direction of friction is opposite to walking i.e., east direction.

As the person does not skid or slip or roll the friction force which is acting on him is static in nature.

Hence, friction force is static and toward east in nature.

Mass of the Body: $6 \, \text{kg}$

Initial Velocity ($u$): $4 \, \text{m/s}$

Final Velocity ($v$): $6 \, \text{m/s}$

Force Applied ($F$): $12 \, \text{N}$

We start by determining the acceleration ($a$) using Newton's second law, defined as:

$ F = m \cdot a $

Substituting the known values:

$ a = \frac{F}{m} = \frac{12}{6} = 2 \, \text{m/s}^2 $

Next, we use the first equation of motion to find the time ($t$) taken for the velocity change:

$ v = u + a \cdot t $

Rearranging for $t$:

$ t = \frac{v - u}{a} = \frac{6 - 4}{2} = 1 \, \text{s} $

To find the displacement ($s$), we use the second equation of motion:

$ s = u \cdot t + \frac{1}{2} a \cdot t^2 $

Substituting the known values:

$ s = 4 \times 1 + \frac{1}{2} \times 2 \times (1)^2 $

$ s = 4 + 1 = 5 \, \text{m} $

Thus, the displacement of the body is $5 \, \text{m}$.

The given situation is shown below

Body loose contact with surface when reaction is zero, thus

$ \begin{aligned} & F \sin 45^{\circ}=m g \\ \Rightarrow \quad & F=\frac{m g}{\sin 45^{\circ}}=\frac{1 \times 10}{\frac{1}{\sqrt{2}}} \text { or } F=10 \sqrt{2} \end{aligned} $

But, $F=\alpha t$

Hence, $\alpha t=10 \sqrt{2}$ or $t=\frac{10 \sqrt{2}}{\alpha}$

Now, acceleration of body, $a=\frac{F}{m} \cos 45^{\circ}$

$ \begin{array}{ll} \Rightarrow & \frac{d v}{d t}=\frac{\alpha t}{m} \cos 45^{\circ} \\ \text { or } & d v=\frac{\alpha}{\sqrt{2} \times 1} \times t \times d t \\ \Rightarrow & d v=t \cdot \frac{\alpha}{\sqrt{2}} \cdot d t \\ \Rightarrow & \int_0^v d v=\frac{\alpha}{\sqrt{2}} \int_0^{10 \sqrt{2}} t \cdot d t \\ \Rightarrow & v-0=\frac{\alpha}{\sqrt{2}}\left(\frac{t^2}{2}\right)_0^{\frac{10 \sqrt{2}}{\alpha}} \\ \Rightarrow & v=\frac{\alpha}{\sqrt{2}} \times \frac{100 \times 2}{2 \times \alpha^2}=\frac{50 \sqrt{2}}{\alpha} \mathrm{~m} / \mathrm{s} \end{array} $

Force applied on block $A, F=10 \mathrm{~N}$

Acceleration produced in block $A$.

$ a=20 \mathrm{~m} / \mathrm{s}^2 $

According to Newton's second equation of motion,

$ \begin{aligned} F & =m_A a \\ m_A & =\frac{F}{a}=\frac{10}{20}=0.5 \mathrm{~kg} \end{aligned} $

Again, block $A$ is kept in contact with block $B$, where $m_B=1.5 \mathrm{~kg}$.

If $a_1$ be the common acceleration in the block system and $R$ be reaction force on block $A$ and $B$, then from free body diagram of block $A$ and $B$,

Equation of motion of block $A$ is

$ \begin{aligned} &\begin{aligned} & & F^{\prime}-R & =m_A \cdot a_1 \\ \Rightarrow & & 20-R & =0.5 a_1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned}\\ &\text { For block } B, R=m_B a_1 \end{aligned} $

$ R=1.5 a_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

From Eqs. (i) and (ii), we have

$ \begin{aligned} & & 20-1.5 a_1 & =0.5 a_1 \\ \Rightarrow & & 20 & =2 a_1 \\ \Rightarrow & & a_1 & =10 \mathrm{~m} / \mathrm{s}^2 \end{aligned} $

∴ From Eqs. (ii),

$ R=1.5 a_1=1.5 \times 10=15 $

Hence, force on block $B$ will be 15 N .

Initial velocity of motor car,

$ u=7 \mathrm{~m} / \mathrm{s} $

Stopping distance, $s=10 \mathrm{~m}$

Final velocity, $v=0 \mathrm{~m} / \mathrm{s}$

If retardation of car is $a \mathrm{~m} / \mathrm{s}^2$, due to applying brakes, then

$ \begin{aligned} a & =\frac{v^2-u^2}{2 s} \\ & =\frac{0^2-7^2}{2 \times 10}=\frac{-4.9}{2} \mathrm{~m} / \mathrm{s}^2 \end{aligned} $

Negative sign indicates that retardation is applied in opposite direction of motion.

$ \text { } \begin{aligned}∴ Resistance,\,\, R & =m a=\frac{w}{9.8} \times \frac{-4.9}{2} \\ & =-\left(\frac{w}{4}\right) \end{aligned} $

The given situation is shown below.

Let's assume the block is stationary only till point $P$. Slope of point $P=\tan \theta$ (where, $\theta=$ angle of repose)

also : $\frac{d y}{d x}=\frac{d}{d x}\left(\frac{x^2}{20}\right)=\frac{x}{10}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

We know that, the block will remain stationary only up till angle of repose.

We know, angle of repose $\theta=\tan ^{-1}\left(\mu_s\right)$

$ \begin{array}{ll} \Rightarrow & \theta=\tan ^{-1}(0.5) \\ \Rightarrow & \tan \theta=0.5 \end{array} $

Now, equating $(d y / d x)$ at $P$ with $\tan \theta$ from Eqs. (i) and (ii), we get

$ \begin{gathered} \frac{d y}{d x}=\tan \theta \\ \Rightarrow \quad \frac{x}{10}=0.5 \Rightarrow x=5 \,\,...(ii)\\ \therefore \quad y=\frac{x^2}{20}=\frac{25}{20}=1.25 \mathrm{~m} \end{gathered} $

According to the question,

$ \begin{aligned} f_1 & =\text { Friction on } 1 \mathrm{~kg}=\mu, N_1=0.4 \times \frac{10}{\sqrt{2}} \mathrm{~N} \\ & =\frac{4}{\sqrt{2}} \mathrm{~N} \\ f_2 & =\text { Friction on } 2 \mathrm{~kg} \\ & =\mu, N_2=0.4 \times \frac{20}{\sqrt{2}}=\frac{8}{\sqrt{2}} \mathrm{~N} \end{aligned} $

Applying Newton's second law along $X$-axis. Since, both are connected, so assuming common acceleration $=a$

$ \begin{aligned} &\begin{aligned} \therefore \quad & \left(\frac{10}{\sqrt{2}}+\frac{20}{\sqrt{2}}\right)-\left(f_1+f_2\right)=1 a+2 a \\ & \frac{30}{\sqrt{2}}-\left(\frac{4}{\sqrt{2}}+\frac{8}{\sqrt{2}}\right)=3 a \\ & \frac{30}{\sqrt{2}}-\frac{12}{\sqrt{2}}=3 a \\ & \frac{18}{\sqrt{2}}=3 a \\ & a=\frac{6}{\sqrt{2}} \end{aligned}\\ &\text { Since, given in question, } a=\alpha \sqrt{2}\\ &\begin{aligned} \therefore & & \alpha \sqrt{2} & =\frac{6}{\sqrt{2}} \\ \Rightarrow & & \alpha & =\frac{6}{2}=3 \end{aligned} \end{aligned} $

Given,

$ \begin{aligned} & m=100 \mathrm{~g}=0.1 \mathrm{~kg} \\ & \Delta t=2 \mathrm{~ms}=2 \times 10^{-3} \mathrm{~s} \end{aligned} $

Initial speed, $u=10 \mathrm{~m} / \mathrm{s}$

Let final speed be $v$.

We know that, area under force-time graph gives impulse or change in momentum.

∴ Area of the triangle $=|\Delta \mathbf{p}|$ (where, $\mathbf{p}=$ momentum)

$ \begin{aligned} & \Rightarrow \quad \frac{1}{2} \times \text { Base × Height }=\left|\mathbf{p}_f-\mathbf{p}_i\right| \\ & \left(\text { where, } \mathbf{p}_f=\right.\text { final momentum and } \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \left.\mathbf{p}_i=\text { initial momentum }\right) \\ & \Rightarrow \frac{1}{2} \times\left(2 \times 10^{-3}\right) \times(20 \times 1000)=0.1 v-0.1 u \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\because 20 \mathrm{kN}=20000 \mathrm{~N}) \end{aligned} $

$ \begin{array}{lc} \Rightarrow & 2 \times 10=0.1(v-10) \quad(\because u=10 \mathrm{~m} / \mathrm{s}) \\ \Rightarrow & v-10=200 \\ \Rightarrow & v=210 \mathrm{~m} / \mathrm{s} \end{array} $

Given, mass of object, $m_2=2 \mathrm{~kg}$ Velocity, $\mathbf{v}=\left(8 t \hat{\mathbf{i}}+3 t^2 \hat{\mathbf{j}}\right) \mathrm{m} / \mathrm{s}$

∴ Acceleration of object is given as

$ \begin{aligned} & \mathbf{a}=\frac{d \mathbf{v}}{d t}=\frac{d}{d t}\left(8 t \hat{\mathbf{i}}+3 t^2 \hat{\mathbf{j}}\right) \\ & \mathbf{a}=8 \hat{\mathbf{i}}+6 t \hat{\mathbf{j}} \mathrm{~ms}^{-2} \end{aligned} $

Hence, according to Newton's second law of motion, force on the object,

$ \Rightarrow \quad \begin{align*} \mathbf{F} & =m \mathbf{a}=2(8 \hat{\mathbf{i}}+6 t \hat{\mathbf{j}}) \\ \mathbf{F} & =16 \hat{\mathbf{i}}+12 t \hat{\mathbf{j}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ |\mathbf{F}| & =\sqrt{16^2+(12 t)^2} \end{align*} $

$ \begin{array}{lc} \Rightarrow & 20=\sqrt{16^2+(12 t)^2} \\ & {[\text { Given, }|\mathbf{F}|=20 \mathrm{~N}]} \\ \Rightarrow & 400=256+144 t^2 \\ \Rightarrow & 144 t^2=144 \Rightarrow t^2=1 \Rightarrow t=1 \mathrm{~s} \end{array} $

Put $t=1 \mathrm{~s}$ in Eq. (i), we get

$ \mathbf{F}=16 \hat{\mathbf{i}}+12 \hat{\mathbf{j}} $

∴ Angle made by force $\mathbf{F}$ relative to positive direction of $X$-axis is given as

$ \begin{aligned} \theta & =\tan ^{-1}\left(\frac{F_Y}{F_X}\right) \\ & =\tan ^{-1}\left(\frac{12}{16}\right)=\tan ^{-1}\left(\frac{3}{4}\right) \end{aligned} $

All forces are resolved in two perpendicular axes ( $X$ and $Y$ ) as shown in the figure.

Since, block of mass $m$ is in equilibrium.

Hence, resolving the forces in $x$-direction, we get

$ \begin{aligned} \left|\mathbf{F}_2\right| \cos \left(60^{\circ}\right)= & \left|\mathbf{F}_1\right| \cos \left(30^{\circ}\right) \\ \left|\mathbf{F}_2\right| \times \frac{1}{2}= & 10 \times \frac{\sqrt{3}}{2} \\ & \left(\because\left|\mathbf{F}_1\right|=10 \mathrm{~N} \text { given }\right) \end{aligned} $

$ \left|\mathrm{F}_2\right|=10 \sqrt{3} \mathrm{~N} $

Again, resolving the force in $y$-direction, we get

$ \begin{aligned} \left|\mathbf{F}_3\right| & =\left|\mathbf{F}_1\right| \sin \left(30^{\circ}\right)+\left|\mathbf{F}_2\right| \sin \left(60^{\circ}\right) \\ & =10 \times \frac{1}{2}+10 \sqrt{3} \times \frac{\sqrt{3}}{2} \\ & =5+15=20 \mathrm{~N} \end{aligned} $

The given situation is shown below

Normal reaction is

$ \mathbf{N}=F \cos \theta=F \cos 30^{\circ} $

Friction force when block is just on the verge of sliding is

$ f=\mu N=\mu F \cos 30^{\circ}=\frac{3}{2} F[\therefore \mu=\sqrt{3}] $

As block is in equilibrium, friction $=$ weight of block $(m g)+$ vertical component of force ( $F \sin \theta$ )

$ \Rightarrow \quad \frac{3}{2} F=3 \times 10+F \times \frac{1}{2} \Rightarrow F=30 \mathrm{~N} $

Given,

Momentum of bullet after firing = 20 $\mathrm{kg} \mathrm{ms}^{-1}$

Velocity $=1000 \mathrm{~ms}^{-1}$

We know that, $p=m v$

$ \begin{aligned} & \Rightarrow m=\frac{p}{v}=\frac{20}{1000} \mathrm{~kg} \\ & \Rightarrow m=\frac{20}{1000} \times 1000 \mathrm{~g} \Rightarrow m=20 \mathrm{~g} \\ & \Rightarrow \quad 20 \leq \frac{u^2 \sin 2 \theta}{g} \leq 24.2 \\ & \Rightarrow \quad 20 \leq \frac{u^2 \times \sin \left(2 \times 15^{\circ}\right)}{10} \leq 24.2 \\ & \Rightarrow \quad 200 \leq u^2 \times \frac{1}{2} \leq 242 \quad\left(\because \sin 30^{\circ}=\frac{1}{2}\right) \\ & \Rightarrow \quad 400 \leq u^2 \leq 484 \\ & \Rightarrow \quad 20 \mathrm{~m} / \mathrm{s} \leq u \leq 22 \mathrm{~m} / \mathrm{s} \end{aligned} $

So, option (a) is best choice.

The given situation is shown in the figure. Resolving applied force, we have following forces on block

$ \text { Given, } \tan \theta=\frac{3}{4}, \sin \theta=\frac{3}{5} \text { and } \cos \theta=\frac{4}{5} $

So, horizontal component,

$ F_1=12 \times \cos \theta=12 \times \frac{4}{5}=\frac{48}{5} \mathrm{~N} $

and vertical component,

$ F_2=12 \times \sin \theta=12 \times \frac{3}{5}=\frac{36}{5} \mathrm{~N} $

So, total applied force on ground is

$ 10+\frac{36}{5}+2 \times 10=37.2 \mathrm{~N} $

So, reaction $N_1$ of ground $=37.2 \mathrm{~N}$ upwards

Also, total force horizontally on wall is

$ F_1=\frac{48}{5}=9.6 \mathrm{~N} $

So, reaction $N_2$ of wall $=9.6 \mathrm{~N}$ towards left