Basic Science - Class 10 General Knowledge - Extra Questions

Horizontal component is always constant
$$\vec{v} = 6\hat{i} + x\hat{j}$$
Apply motion equation
$$v = u + at$$
$$3 = u - 10 \times 0.5$$
$$u = 3 + 5 = 8\,m/s = x$$
$$Ans - (6\hat{i} + 8\hat{j}) \,m/s$$
total
times = $$\dfrac{2}{g} \times (y-component)$$
$$= \dfrac{2}{10} \times 8 = 1.6\,sec$$

For calculating the total number of protons in a polyatomic ion, one must add the atomic number of all atoms in a polyatomic ion.

$$v = t^{2} - t$$
$$\frac{dv} {dt} = 2t - 1$$
$$2t - 1 = 0$$
$$t = \frac{1} {2}$$

According to equation

$$8.0\ Kg. m/s$$
Explanation:
Since the ball bounces back without losing kinetic energy, its speed , and hence the magnitude of its momentum remains the same after teh collision, only the direction reverses.
The change in momentum of the ball is $$8.0\ Kg.m/s$$
Since the force that the ball exerts on the wall is equal and opposite to that exerted on it by the wall, the impulse exerted on the wall is equal in magnitude (through opposite in direction) to that exerted on the ball.

Matter can neither be created nor destroyed. This is the law of conservation of matter (mass).

Time taken by A to travel $$30 \sin 45^{\circ} $$ 

$$t = \dfrac{30 \sin 45^{\circ}}{10} $$ 

$$ = \dfrac{3}{\sqrt{2}} $$ 

$$ = \dfrac{3}{2} \sqrt{2} $$ hr.

$$V_{B} \cos \theta = \dfrac{30 \sin 45^{\circ}}{\dfrac{3}{2} \sqrt{2} hr} $$ 

$$ \cos \theta = \dfrac{2}{3} $$ 

$$ \boxed{\theta = \cos^{-1} \left  (\dfrac{2}{3}  \right  )} $$

Relative velocity in horizontal direction $$ = V_{B} \sin \theta - V_{A} $$ 
$$ = (15 \sin \theta -10) $$ 

$$ \Rightarrow (15 \sin \theta - 10) = \dfrac{30 \sin 95^{\circ}}{t} $$   .............(i) 

In vertical direction
$$15 \cos \theta - 0 = \dfrac{30 \cos 145^{\circ}}{t} $$  ..........(ii) 

$$\Rightarrow 15 \cos \theta = 15 \sin \theta - 10 $$ 

$$\Rightarrow 10 = 15 (\sin \theta - \cos \theta) $$ 

$$\Rightarrow \sin \theta - \cos \theta = \dfrac{2}{3} $$  

Square both sides
$$(\sin \theta - \cos \theta)^{2} = \left  (\dfrac{2}{3}  \right  )^{2} $$ 

$$ \Rightarrow \sin^{2} \theta + \cos^{2} \theta - 2 \sin \theta \cos \theta = \dfrac{4}{9} $$ 

$$ \Rightarrow 1 - \dfrac{4}{9} = 2 \sin \theta \cos \theta $$ 

$$\Rightarrow \dfrac{5}{9} = \sin 2 \theta $$ 

$$ \Rightarrow \theta = \dfrac{1}{2} \sin^{-} \dfrac{5}{9} = \left  (\dfrac{33.7}{2}  \right)^{\circ} $$ 

$$ \approx 17^{\circ} $$

Its focal length is long, so it gives a wider field of view.

He bends in order to keep to the centre of gravity of body within the base. This will save him from falling.

Accumulated steam in the cooker which cannot escape increases pressure. So the temperature will also increases.

Water does not allow the flames to come in contact with air to get oxygen.

$$F - f = mg\sin \theta$$
$$F = mg\sin \theta \ + \ f$$
$$= 30\times 10 \times \dfrac {1}{2} + 150$$
$$= 300\ N$$
$$F'\cos 30 = F$$
$$F' = \dfrac {F}{\cos 30} = \dfrac {300}{\dfrac {\sqrt {3}}{2}}$$.

$$\Delta L=I{w}_{2}-I{w}_{1}$$
$$ =2\pi I({f}_{2}-{f}_{1})$$
$$=2\pi \times 6.66(0.5-1)$$
$$=21.21kg$$ $${m}^{2}{s}^{4}$$

$$t_{ABCDE} = 8s $$ 
$$t_{ABC} = \dfrac{8}{2} = 4 $$ 
$$t_{BC} = \dfrac{4}{2} = 2 $$ 
or $$ [t_{AB} = 2] $$

Now total time $$ = 8\ sec $$ 

$$ u = (-v) + gt $$     $$ [V = u + at , v = u , v = -u , a = g] $$ 

$$ 2u = gt $$ 

$$ 2 u - 10 \times 8 $$ 
$$u = 40 $$ 

Now $$ h = ut - \dfrac{1}{2} gt^{2} $$ 
$$ h = 40 \times 2 - \dfrac{1}{2} \times 10 \times 4 $$ 
$$ t = 2 $$ 
$$ h = 60 $$ 

nitial velocity of bullet $$v_{1}=50\sqrt{7}m/s$$

$$V_{L} = \sqrt{5 Rg} $$ 
$$ = \sqrt{5 \times1 \times 9.8} $$ 
$$ = \sqrt{49} $$ 
$$ =7\ m/s $$

$$V_{H} = \sqrt{rg} $$ 
$$ = \sqrt{1 \times 9.8} $$ 
$$ = 3.1\ m/s $$

$${V}^{2}=mx$$
$${V}^{2}=\tan {37}^{o})x$$
$${V}^{2}=\cfrac{3}{4}x$$
$$\Rightarrow$$ $$2V\cfrac{dv}{dt}=\cfrac{3}{4}.\cfrac{dx}{dt}$$
$$2V\cfrac{dv}{dt}=\cfrac{3}{4}\times V$$
$$\Rightarrow$$ $$\cfrac{dv}{dt}=a=\cfrac{3}{8}$$

We know that moment of inertia of a wheel is given by

$$T = 2 \pi \sqrt{\frac{L}{g}}$$
                          $$L = 2R$$
  $$= 2 \pi \sqrt{\frac{2R}{g}}$$

Man 'A' moves in north direction with speed 10m/s.

Speed of thief, $$V_{th}$$ = 10 m/s
Speed of police, $$V_{pol}$$ = 12 m/s
Distance, s = 80 m
Relative Speed = $$V_{pol} - V_{th}$$
= 2 m/s
Time taken to catch = $$\dfrac{80}{2}$$
= 40 sec

Average velocity $$ = \dfrac{1 + 2 + 3 + 4 + 5}{5} m/s $$ 

$$ = \dfrac{15}{5}\ m/s $$ 

$$ =3\ m/s $$

$$m_{1} = \rho V_{1} = M $$ 
$$m_{2} = \rho V_{2} = 8M $$ 
$$m_{3} = \rho V_{3} = 27 M $$ 
$$m_{4} = \rho V_{4} = 64 M $$ 

$$ y_{1} = \dfrac{10}{2} = 5 $$
$$ y_{2} = \dfrac{20}{2} = 10$$
$$ y_{3} = \dfrac{30}{2} = 15 $$ 
$$ y_{4} = \dfrac{40}{2} = 20 $$ 

$$V_{1} = (10)^{3} = 1000 $$
$$V_{2} = (20)^{3} = 8000 $$ 
$$V_{3} = (30)^{3} = 27000 $$ 
$$V_{4} = (40)^{3} = 64000 $$ 

$$y_{CM} = \dfrac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3} + m_{4} y_{4}}{m_{1} + m_{2} + m_{3} + m_{4}} $$

$$y_{CM} = \dfrac{M \times 5 + 8M \times 10 + 27M \times 15 + 64 M \times 20}{M + 8M + 27M + 64M} $$ 

$$y_{CM } = 17.7\ cm $$ 

$$x_{1} = \dfrac{10}{2} = 5\ cm $$ 
$$x_{2} = 10 + \dfrac{20}{2} = 20\ cm $$ 
$$x_{3} = 10 + 20 + \dfrac{30}{2} = 45\ cm$$ 
$$x_{4} = 10 + 20 + 30 +\dfrac{40}{2} = 80\ cm $$ 

$$x_{CM} = \dfrac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}} $$ 

$$x_{CM} = \dfrac{m \times 5 + 8m \times 20 + 27 m \times 45 + 64m \times 80}{m + 8m + 27m + 64m} $$ 

$$x_{CM} = 65\ cm $$

mi = $$9.1 \times 10^{-31} $$ kg
charge = $$1.6 \times 10^{-19} $$C
radius = 0.5 m, velocity = $$10^{6} $$m/s
now $$ r = \frac{mv} {qB}$$ So, $$0.5 =  \frac{(9.1 \times 10^{-31} ) (10^{6})} {(1.6 \times 10^{-19} ) (B)}$$
$$\Rightarrow B = \frac{9.1 \times 10^{-25}} {1.6 \times 10^{-19} \times 0.5}$$
$$\frac {9.1 \times 10^{-6}} {0.8}$$
$$= 11.375 \times 10^{-6}$$ Tesla

$$w = 2\pi f$$

Answer 
$$N_{1} F - 3_{a} = F - \dfrac{mF}{7m} $$
$$ N_{1} = \dfrac{7F - F }{7} = \dfrac{6F}{7}$$
$$ F_{1} N_{1} = ma , N_{1} -N_{2} = 2na , N_{2} = 4ma $$
$$ a = \dfrac{F}{m + 2m + 4m} = \dfrac{F}{7m}$$
$$N_{2} = 4 m \times \dfrac{F}{7m} = \dfrac{4F}{7}$$

$$F = mg\,sin\theta - N$$
when ball loses contact, $$N = 0$$
$$\Rightarrow F = mg \,sin\theta$$
$$\because$$ Ball is moving in circular path
$$F = \dfrac{mv^2}{R}$$
$$\Rightarrow \dfrac{mv^2}{R} = mg\,sin\theta$$
$$\Rightarrow v^2 = Rg\,sin\theta$$
From conservation of energy
$$\Rightarrow mgR = \dfrac{1}{2}mv^2 + mgh$$
$$\Rightarrow Rg = \dfrac{1}{2}Rg \,sin\theta + gh$$
$$\Rightarrow h = R - \dfrac{1}{2}R sin\theta$$
$$\because sin\theta = \dfrac{h}{R}$$
$$\Rightarrow h = R - \dfrac{1}{2}R\dfrac{h}{R}$$
$$\Rightarrow \dfrac{3h}{2} = R$$
$$\Rightarrow h = \dfrac{2R}{3}$$

For just to reach the top velocity at top = 0
Let minimum velocity  = v
$$KEi = \dfrac{mv^2}{2}$$
$$PEi = 0$$
$$Ei = \dfrac{mv^2}{2}$$ ........(i)
at top
$$KE_f = 0$$
$$PE_f = mgh = mg(2R)$$
$$E_f = 2mgR$$ ........(ii)
using conservation of energy, $$E_i = E_f$$
$$\dfrac{mv^2}{2} = 2mgR$$
$$v^2 = 4gR$$
$$v = 2\sqrt{gR}$$
when ball is at height h initially
$$E_f = \dfrac{mv^2}{2} + mgh$$
at top
$$E_f = 0 + mg(2R) = 2mgR$$
$$E_i = E_f$$
$$\dfrac{mv^2}{2} + mgh = 2mgR$$
$$\dfrac{v^2}{2} = 2gR - h$$
$$v^2 = (4gR - 2h)$$
$$v = \sqrt{(4gR - 2h)} = \sqrt{2(2gR - h)}$$

Let on object on moving in uniform acceleration
u = initial velocity of object
v = final velocity of object
a = uniform velocity of object
Let object reach point B other time (t)
Slope = Acceleration (a) = $$\dfrac{\text{change in velocity}}{time}$$
Change in velocity $$AB = \bar{v} - \bar{u}$$
time $$AD = t$$
$$\therefore a = \dfrac{v - u}{t}$$ or $$v = u + at$$

$$V(t) = 230 \sin (314t - 30) $$ 
$$ = 230 \sin \left  (314 t - \dfrac{\pi}{6}  \right  ) $$ 

Now using KVL 
$$230 \sin \left  (314 t - \dfrac{\pi}{6} \right  ) - L \dfrac{di}{dt} = 0 $$ 

$$ \dfrac{di}{dt} = \dfrac{230}{2.27} \sin \left (314 t - \dfrac{\pi}{6}  \right  ) $$ 

$$ i = \dfrac{-230}{314 \times 2.27} \sin \left  (314t- \dfrac{\pi}{6}  \right  ) $$ 

$$ i = -0.323 \sin \left  (314 t - \dfrac{\pi}{6}  \right  ) $$

from $$2$$ to $$9h/16$$
$$v^2 = u^2 + 2gs$$
$$v = \sqrt{2gs}$$
$$= \sqrt{2 \times g \times \dfrac{9h}{16}}$$
$$= \sqrt{\dfrac{18gh}{16}}$$ .......(i)
from $$\dfrac{7h}{16}$$ to ground
$$\dfrac{7h}{16} = \sqrt{\dfrac{18gh}{16}} (1) + \dfrac{1}{2} \times g \times (1)^2$$
$$\dfrac{7h}{16} = \sqrt{\dfrac{18g}{16}} \sqrt{h} + 5$$

We have 
$$-mgx + \dfrac{1}{2} kx^{2} = 0 $$ 

$$x = \dfrac{2mg}{k} $$

Net gravitational field intensity
$$=\frac{G(m_2-m_1)}{r^2}$$
$$=\frac{G(1000-100)}{(\frac{1}{2})^2}$$
$$=6.67*4*900*10^{-4}$$
$$=2.4*10^{-8}Nkg^{-1}$$
Direction along $$1000kg$$

$$ \tau = I \alpha $$ 
$$I = Mk^{2} $$

$$ \tau = Mk^{2} \alpha $$ 

$$ k = \sqrt{\dfrac{\tau}{M \alpha}} $$ 

$$ = \dfrac{400}{40 \times 20} $$ 

$$ = \sqrt{\dfrac{1}{2}} $$ 

$$ = \dfrac{1}{\sqrt{2}} m $$

$$\dfrac{\text{change in momentum}}{time} = Force$$
$$momentum = mv$$
$$Force = \dfrac{n×m×v}{t} = \dfrac{40×0.035×4}{1}= 5.6\,N$$

No, atoms of two different elements can not have the same number of protons.

Atoms from two different elements may have the same number of neutrons, but never the same number of protons. The number of protons is unique to the element and represents the atomic number.

$$F_1 = F_2 = \dfrac{K\,q\cdot q_2}{OA^2}$$
$$= \dfrac{Kq_1 \cdot q_2}{\left ( \dfrac{d}{2} \right )^2 + x^2}$$
$$F_{net} = 2F_1\,cos\theta$$
$$= \dfrac{Kq_1q_2}{x^2 + (\dfrac{d}{2})^2} = \left ( \dfrac{OC}{OA} \right )$$
$$= \dfrac{Kq_1q_2}{x^2 + (\dfrac{d}{2})^2} = \dfrac{x}{\sqrt{x^2 + (\dfrac{d}{2})^2}}$$
$$y = F_{net} = Kq_1q_2 \left ( \dfrac{x}{\left (x^2 + (\dfrac{d}{2})^2 \right )^{3/2}} \right )$$
Put $$\dfrac{dy}{dx} = 0$$
$$\dfrac{\left (x^2 + (\dfrac{d}{2})^2 \right )^{3/2} - x \left (x^2 + (\dfrac{d}{2})^2)^{1/2} (2x) \right )}{\left (x^2 + (\dfrac{d}{2})^2 \right )^3}$$
$$= \left (x^2 + (\dfrac{d}{2})^2 \right ) - 3x^2 = 0$$
$$2x^2 = \left ( \dfrac{d}{2} \right )^2$$
$$x^2 = \left ( \dfrac{d}{2} \right )^2 \times \dfrac{1}{2}$$
$$x = \dfrac{d}{2\sqrt 2}$$

As we know efflux vel. of water is $$ v = \sqrt{2gh} $$ 

Time to reach ground level by water is $$t$$ 

$$t = \sqrt{\dfrac{2 (H - h)}{g}} $$ 

Horizontal range on ground $$ = v.t $$ 
$$ = 2 \sqrt{h (H - h)} $$ 

For x to be max $$ \dfrac{dx}{dh} = 0 $$ 

$$ 0 = \dfrac{1}{2 \sqrt{H (H - h)}} \times (H - 2h) $$ 

$$H = 2h $$ 

$$ \boxed{ h = \dfrac{H}{2}}$$

Answer
[C] =  [t]
$$ [C]  = [M^{o}L^{o}T] $$
$$ \left [ \dfrac{b}{t + c} \right ]   =  [v] $$
$$ [b] =  [v] [ t + c] $$
$$  =  [LT^{-1}] [T] $$
$$  = [M^{o}L^{1}T^{o}] $$
$$[v] = [at]$$
$$[LT^{-1}] = [a] [T]$$
$$\therefore a = [LT^{-2}]$$
$$[t] = [c] = [T]$$
$$[B] = [v]$$
$$[b] = [v][t]$$
$$= [LT^{-1}] [T]$$
$$= [L]$$

Work done =  area under f-x curve so here are 

$$=ma$$

$$Torque\, T=\bar{r}\times \bar{F}$$

About O,

Norma  force external by the wall
and $$ N  =  m g \cos  \theta $$

From conservation of energy

$$V_{e}=\sqrt{\dfrac{2GM}{R}}$$
$$V_{other}=\sqrt{\dfrac{2GM'}{R'}}$$
$$\dfrac{V_e}{V_{other}}\sqrt{\dfrac{2GM}{R}\times \dfrac{R'}{2GM'}}=\sqrt{\dfrac{MR'}{M'R}}$$
$$=\sqrt{\dfrac{R}{4R}} \left(R'=\dfrac{R}{4}, M'=M\right)$$
$$\dfrac{V_{e}}{V_{o}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\Rightarrow V_o =2\ V_o$$
$$V_{o}=22.4\ m/s$$

Newton's First Law of Motion
An object at rest will remain at rest and an object in motion will remain in motion at constant velocity unless acted upon by an unbalanced force.
Newton's Second Law of Motion
The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. $$Force = Mass\ X$$ Acceleration.
Newton's Third Law of Motion
Every action, there is an equal and opposite reaction.

Answer
The value of a depth  is given by
$$g_{d} =  g \left ( 1 - \dfrac{d}{R} \right ) $$
Here $$ g_{d} = \dfrac{9}{4} $$
$$ \dfrac{9}{4} =  g \left ( 1 - \dfrac{d}{R} \right ) $$
$$ \dfrac{d}{R} = \dfrac{3}{4} $$
$$ d  =  0.75  \times 6400 = 4800 km $$

The Balmer series is a set of emission lines of the Hydrogen spectrum in which the electron in an excited state $$\mathrm{n\geq 3}$$ transits to $$\mathrm{n = 2}$$.

Answer
121.6 nm
Explanation :
$$ \dfrac{1}{\lambda } =  R  \left ( \dfrac{1}{(n_{1})^{2}} - \dfrac{1}{(n_{2})^{2}} \right ) .Z^{2}$$
where,
R = Rydbergs constant (Also written is RH)Z = atomic number
Since the question is asking for 1st line of Lyman series therefore $$ n_{1} =  1 $$
$$ n_{2} =  2 $$
since the electron is de -exited from 1(st) exited state (i.e n=2) to ground state (i.e n=1) for first line of Lyman series.
Therefore plugging in the values
$$ \dfrac{1}{\lambda } = R = \left ( \dfrac{1}{(1)^{2}} - \dfrac{1}{(2)^{2}} \right ).1^{2} $$
Since the atomic number of Hydrogen is 1.
By doing the math, we get the wavelength as
$$ \lambda  = \dfrac{4}{3}.912A $$ since $$\dfrac{1}{R} =  912 A $$ 
therefore  $$ \lambda  =  1216 A $$
or$$  \lambda  =  121. 6nm $$

Answer
Someone has asked for a real life example of negative displacement, so here is one below.
Explanation:
Displacement can be negative because it defines a change in position of an object while carefully monitoring its direction.
For our example the object we will consider is you. If you leave your present position and walk down a hallway to get a drink and you walk 6m to where the drink is located, your displacement is 6m in the h (hallway) direction. We will take this first trip as the positive direction
Then you walk back towards your original location in the h opposite direction. But Surien calls you to her office which is 15 m further past your office. It is still in the direction h opposite the drink location.
Your displacement is now 6 m 6 m 15 m = 15 m in the h direction. Notice both displacements had to be subtracted from the +h of your first trip because both were in the h direction from it.
Then you return to your original location.Now your total displacement is:
6 m 6 m 15 m + 15 m = 0 m in the h direction.
Because of positive and negative displacement vectors, you have essentially gone nowhere. You can verify this by looking around. You are back at your original location.
While accomplishing this feat, you have managed to cover a distance of  6m + 6 m + 15 m + 15 m = 42 m
Time for another drink....

In a stronger gravitational field a given mass will have a
larger weight. 

Be An object's weight is given by this equation: $$w=m g$$

Gravitational field strength at a distance, $$r,$$ from a
centre of mass is given by this equation:

$$g=\frac{G M}{r^{2}}$$

If an object is nearer a centre of mass then $$r$$ will be
smaller so g will be larger. In that case the weight of the object will be
larger. If an object is moved from a distance R away from one centre of mass
and placed a distance R away from the centre of a smaller mass then g will be
smaller. In that case the weight of the object will be smaller.

 A practical example: if an astronaut on earth has a weight
of $$833 \mathrm{N},$$ on the moon he/she will have a weight of $$136
\mathrm{N}$$.

$$\cup$$ sing $$m=\frac{w}{g}$$ the mass of the astronaut is
$$\frac{833}{9.8}=85 \mathrm{kg}$$

So the weight on the moon $$\left(g_{M}=1.6
\mathrm{Nkg}^{-1}\right)$$ is:

$$w_{M}=m g_{M}=85 \times 1.6=136 N$$

The law of conservation of matter and energy states that matter is neither created nor destroyed but conserved.

$$31.6 m s^{-1}$$

This solution assumes an incompressible fluid, so the volumetric flow rate is constant at all points in the hose and nozzle (this is because the density is constant and the mass flow rate must be the same at all points). In order to calculate the of velocity of the water in the

nozzle we need to know the cross-sectional area of the nozzle. $$A_{N}=\dfrac{\pi D_{N}^{2}}{4}=\dfrac{\pi(0.022)^{2}}{4}=3.8 \cdot 10^{-4} \mathrm{m}^{2}$$

Now we can use the cross-sectional area with the volumetric flow rate to calculate the velocity of the water in the nozzle:

$$\dfrac{d V}{d t}=A \dfrac{d s}{d t}$$

Where $$\dfrac{d V}{d t}$$ is the volumetric flow rate and $$\dfrac{d
s}{d t}$$ is the rate of change of displacement, i.e. the velocity.

$$v=\dfrac{(d V) /(d t)}{A}=\dfrac{0.012}{3.8 \cdot 10^{-4}}=31.6 \mathrm{ms}^{-1}$$

Positively charged ions are particles which have a greater number of protons than electrons.

Nutation is an assymetrical movement seen in plants due to change in turgor pressure,where the tendrils show spiral movements surrounding a rigid support.

Answer:
12 N  in the direction opposite to electric field.
Explanation:
Force on a charged particle q due to electric field E is given by
$$\vec{F} = q\vec{E}$$
$$|F| = -0.06 C \times 200 \dfrac{NC} = -12 N$$
Here negative indicates that force is in the direction opposite to electric field.

$$m v = 2m v' $$ 
$$v' = \dfrac{v}{2} = 3\ m/s $$ 

$$mv = (2m + m) v'' $$ 
$$v'' = \dfrac{mv}{2m + m} = \dfrac{1 \times v}{6} = 1\ m/s $$ 

First after collision
Convention of energy

$$ PE + \dfrac{1}{2} \times 2 \times (3)^{2} = E_{i} = 9\ J $$ 

$$ \Rightarrow \dfrac{1}{2} \times 6 \times 1 +2g \times l (1 - \cos \theta) = 9 $$ 

$$1 - \cos \theta = \dfrac{6}{2g l} $$ 

$$1 - \cos \theta = \dfrac{6}{2 \times 10 \times 1.5} $$ 

$$ 1 - \cos \theta = \dfrac{1}{5} $$ 

$$ \cos \theta = \dfrac{4}{5} $$ 

$$\theta = 37^{\circ} $$ 

Let $$R$$ = natural length of spring 
$$OA = \sqrt{2} R $$ 
$$x = \sqrt{2} R - R = R (\sqrt{2} - 1) $$

Acc. to conservation of energy 
$$E_{B} = E_{A} $$ 
$$ \dfrac{1}{2} k (2R - R)^{2} + mg 2R = \dfrac{1}{2} kx^{2} + mgR + \dfrac{1}{2} m $$ 

$$ \dfrac{1}{2} kR^{2} + 2 mg R = \dfrac{1}{2} k [R (\sqrt{2} - 1)]^{2} + mgR + \dfrac{1}{2} mv^{2} $$ 

Put $$ k = \dfrac{1}{\sqrt{2} - 1} $$ 

$$ \dfrac{R^{2}}{2 (\sqrt{2} - 1)} + mgR - \dfrac{1}{2 (\sqrt{2} - 1)} R^{2} (\sqrt{2} - 1)^{2} = \dfrac{1}{2} mv^{2} $$ 

$$v^{2} = \dfrac{2}{m} \begin{vmatrix} mgR + \dfrac{\sqrt{2} R^{2}}{\sqrt{2} - 1}  \end{vmatrix} $$ 

$$ v = \sqrt{2 \left  (Rg + \dfrac{\sqrt{2} R^{2}}{m (\sqrt{2} - 1)}  \right  )} $$ 

Smaller block tend to slip down under the force that is the component of its weight along the plane = mg sin. 
In order to prevent the block from slipping, this block along with the triangular block needs to be accelerated by the system so that the components of pseudo force on the block balances the component of weight which is trying to move the block.