Physics NCERT Exemplar Solutions Class 12th Chapter Four Moving Charges and Magnetism

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Answer- b

Explanation – ler r makes an angle $?$ with positive x axis component if r along x-axis.

So  r_x= |r|cos $?$

(r_x)_maximum= |r| (cos $?$ )_maximum

= |r|cos $0$ = r

$?$ =0⁰

R is along positive x-axis.

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Explanation – in biot savart law, magnetic field B is parallel to idlr and idl have its direction along the direction of flow of current.

Here, for the direction of magnetic field at O₂, due to wire carrying I₁ current is B| parallel idl $*$ r or I $*$ k, but it is -j

So the direction at O₂ is along y-direction.

The direction of magnetic force exerted at O2 because of the wire along x-axis

F= ilB=j (-j)=0

So the force is zero.

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Explanation- force on charge particle due to both electric and magnetic field . now if electric force and magnetic field is zero then electric field and magnetic field is also zero.

Magnetic field is only responsible for constant velocity of charge particle if B=0 then E =0 if B is not equal to zero then electric field is also not equal to zero

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Answer – (a)

Explanation- in biot savarts law magnetic field is perpendicular to v

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Explanation- i_G (G) = (I₁-I_G) (S₁+S₂+S₃) for I₁= 10mA

i_G (G+S₁) = (I₂-I_G) (S₂+S₃) for I₂= 100mA

i_G (G+S₁+S₂) = (I₃-I_G) (S₃) for I₃= 1A

S₁= 1W, S₂= 0,1W and S₃= 0.01W

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Explanation- In Biot-Savart’s law, magnetic field B is parallel (II) to id × r and idl have its direction along the direction of flow of current.

Here, for the direction of magnetic field, At dl2, located at (0, R,0) due to wire d1 is given by B I | dl × r or I ×j (because point (0, R,0)lies on y-axis), but I × j= k

So, the direction of magnetic field at d2 is along z-direction.

The direction of magnetic force exerted at d2 because of the first wire along the x-axis

F =i (l $\times$ B) i.e., F| (i $\times k$ ) or along− j direction.

Therefore, force due to dl1 on dl2 is non-zero.

Now, for the direction of magnetic field, At d1, located at (0,0,0) due to wire d2 is given by

B I | dl × r or j ×−j (because origin lies on y-direction w.r.t. point (0, R,0).), but j ×−j =0.

So, the magnetic field at d1 does not exist.

Force due to dl2 on dl1 is zero.

So, magnetic forces do not obey Newton's third law.

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Explanation – for equilibrium balance net torque should be zero

Mgl= Wcoill

500gl = Wcoill

Wcoil= 500 $*$ 9.8N

Taking moment of force about mid point then magnetic field

Mgl+mgl=Wcoill+IBlsin90

Mgl=BILl

M= $\frac{0.2*4.9*1*{10}^{-2}}{9.8}=10$ -3kg= 1g

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Explanation- the thicker wire has a resistance R, then the other wire has a resistance @R as the wires are of the same material but different area

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Explanation- Since, B is along the x-axis, for a circular orbit the momenta of the two particles are in the y-z plane. Let P1 and P2 be the momentum of the electron and positron, respectively. Both traverse a circle of radius R of opposite sense. Let P1 make an angle? with the y-axis P2 must make the same angle.

The centres of the respective circles must be perpendicular to the momenta and at a distance R. Let the centre of the electron be at Ce and of the positron at Cp . The coordinates of Ce is

Ce= (0, -Rsin $\mathrm{\theta },\mathrm{R}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{\theta }$ )

Cp= (0, -Rsin $\mathrm{\theta },-\frac{3}{2}\mathrm{R}-\mathrm{R}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{\theta }$ )

The circles of the two shall not overlap if the distance between the two centers are

Greater than 2R.

Let d be the distance between Cp and Ce.

Let d be the distance between Cp and Ce.

So d²= (2Rsin $\theta$ )²+ ( $\frac{3}{2}R-2Rcos\theta$ )²

= 4R²sin² $\theta$ + $\frac{9}{4}R$ ²-6R²cos $\theta$ +4R²cos² $\theta$

= 4R²+ $\frac{9}{4}R$ ²-6R²cos $\theta$

Since d has to be greater than 2R

So d²> 4R²

4R²+ $\frac{9}{4}R$ ²-6R²cos $\theta$ > 4R²

Cos $\theta$ <3/8

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Explanation- as we know magnetic moment = nIA

For equilateral triangle M= nIA= 4I ( $\frac{\sqrt[]{3}}{4}{a}^2$ )

M= Ia²$\sqrt[]{3}$

For square, n=3 so total length of wire is 12a

M= nIA= 3I (a²) = 3Ia²

For regular hexagon of side a, n=2 so total length = 12a

M= nIA=2 ( $\frac{6\sqrt[]{3}}{4}{a}^2$ )= 3 $\sqrt[]{3}$ a²I

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Explanation- B (z) points in the same direction on z-axis and hence, J (L) isa monotonically function of L so B.dl = B.dl as cos 0 = 1

(b) $\int B.dl={\mu }_{o}I$ but when L $\to$ $\infty$

So B $\propto$ 1/r³

(c) the magnetic field due to a circular current carrying loop of radius in the xy plane with centre at origin at any point lying at a distance of from origin.

B= $\frac{{\mu }_{0}I{R}^2}{2{(Z^2+R^2)}^{3/2}}$

Z=Rtan $\theta$

dz=Rsec² $\theta d\theta$

${\int }_{-\infty }^{\infty }{B}_{z}dz=\frac{{\mu }_{0}I}{2}{\int }_{-\pi /2}^{\pi /2}cos\theta d\theta$ = ${\mu }_{0}I$

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Explanation- (a) Suppose the five wires A, B, C, D and E be perpendicular to the plane of paper at locations as shown in figure.

Thus, magnetic field induction due to five wires will be represented by various sides of a

closed pentagon in one order, lying in the plane of paper. So, its value is zero.

(b) Since, the vector sum of magnetic field produced by each wire at O is equal to 0.

Therefore, magnetic induction produced by one current carrying wire is equal in

magnitude of resultant of four wires and opposite in direction.

Therefore, the field if current in one of the wires (say A) is switched off is $\frac{{\mu }_{0}I}{2\pi r}$ perpendicular to AO towards left.

(c) if current in wire A is reversed, then total magnetic field induction at O= magnetic field induction due to wire A + magnetic field induction due to wrie B, C, D and E

B= $\frac{{\mu }_{0}I}{2\pi r}$

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Explanation- dW=F.dl=0

As dl = vdt

dW= Fvdt

dW= f.v=0

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Explanation- yes, the magnetic force differ from inertial frame to frame. The magnetic force is frame dependent.

The net acceleration which comes into existing out of this is however, frame independent for inertial frames.

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Explanation – here, the condition of magnetic resonance is violated.

When the frequency of radio frequency field were doubled, the time period of the radio frequency field were halved. Therefore, the duration in which particle completes half revolution inside the dees, radio frequency completes the cycle.

Hence, particle will accelerate and decelerate alternatively. So the radius of path in the dees will remain same.

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Explanation- general formula for magnetic field at center is

B= $\frac{{\mu }_{0}I}{4\pi r}\left(\theta \right)$

B1= $\frac{{\mu }_{0}I}{4\pi r}\left(\frac{\pi }{2}\right)=\frac{{\mu }_{0}I}{8r}$ k for xy plane

B2= $\frac{{\mu }_{0}I}{8r}$ i for zy plane

B3= $\frac{{\mu }_{0}I}{8r}$ j for xz plane

After joining together B= $\frac{{\mu }_{0}I}{8\pi r}$  (i+j+k)

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Explanation- for a charge particle to move in circular path

Mv²/r = qvB

v/r=w= bq/m

w= v/r = $\frac{L{T}^{-1}}{L}$ = T-1

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Explanation- considering magnetic field along B=Bk and electron enters with a velocity v=vi into a cubical region

The force on electron using Lorentz force = -e (vi $*$ Bk)= evBi

So the force F= -eE₀ k accelerates along z axis which in turn increases the radius of circular path and hence particle transversed on spiral path.

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Explanation- applying expression in different situation

For i_G (G+R₁) = 2 , for 2V range

i_G (G+R₁+R₂) = 20 , for 20V range

i_G (G+R₁+R₂+R₃) = 2 , for 200V range

on solving we get R₁= 1990ohm, R₂= 18kohm, R₃= 180k ohm

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Explanation- magnetic field produced by a long straight current carrying wire is

B= $\frac{{\mu }_{0}I}{2\pi r}$ = $\frac{{\mu }_{0}I}{2\pi h}$ , for h height distance

Magnetic force on small conductor is F= ilBsin $\theta$ = ilB

Also force, F =mg = $\frac{{\mu }_0{I}^2}{2\pi r}$ l

h= $\frac{4\pi \times {10}^{-7}\times 25\times 25\times 1}{2\pi \times 2.5\times {10}^{-3}\times 9.8}$ = 51 $\times$ 10^-4cm

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Answer – (d)

Explanation- it is determined by its pitch

Pitch = $\frac{2\pi mvcos\theta }{Bq}$ = constant

So charge by mass ratio is also constant

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Answer- (a)

Explanation-  magnetic moment, M = niA, when we decrease the radius by half its area is decrease by ¼ so the magnetic moment also.so magnetic moment diminishes.

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Answer- (d)

Explanation- when electron move in the direction of velocity along the axis of current then angle is 180 which makes force according to the relation F=qvbsin $?$ . so electron will continue move with uniform velocity along axis of solenoid.

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Answer – (a)

Explanation – when a charge particle enter in cyclotron it will always accelerated with the help of oscillating electric field. But it will move with the same speed with the help of magnetic field.

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Answer. (d)

Explanation- W=MBcos $\theta$ , as the loop make 30 degree angle but angle made by axis of the loop with the direction of magnetic field, therefore work done is zero.

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Answer – (a)

Explanation- gyro magnetic ratio does not depend upon which electron is rotating.

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Answer- (b, d)

Explanation- F=IlB, the direction of force is given by fleming's left hand rule and F is perpendicular to the direction of magnetic field. So work done is zero.

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Answer- (b, c)

Explanation- if current is in opposite sense the? Bdl = 0, also there exist a point where b and dl is perpendicular.

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Answer- (b, c, d)

Explanation- magnetic forces on both the particles cause equal acceleration because they both have equal velocity

Also they both gain or loose energy because direction of force in both cases is opposite.

And there is no change in center of mass of particles so it is determined by B alone.

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Explanation- for a charge particle to move in circular path

Mv²/r = qvB

v/r=w= bq/m

w= v/r = $\frac{L{T}^{-1}}{L}$ = T-1

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Explanation- vsin $\theta$ = vertical component

V_x= horizontal component of velocity =vcos $\theta$ = constant

V_y = vertical component of velocity =vsin $\theta$

Velocity will always be tangential to the curve in the direction of motion and acceleration is always vertically downward and is equal to g.

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Explanation- at point B it will gain the same speed u and after that speed increases and will be maximum just before reaching point c.

During journey from O to A speed decreases and will be maximum at point A and acceleration is always constant.

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Explanation- if the football is kicked into the air vertically upwards. Acceleration of football will always vertically downwards and equal to acceleration due to gravity. But when it reaches the highest point its velocity is zero and acceleration is retarding.

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Explanation – the direction of (B $*C$ ) will be perpendicular the plane containing B and B by right hand rule. A $*(B*C)$ will lie in the plane of B and C and is perpendicular to vector A.

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Explanation – The path of the ball observed by a boy standing on the footpath is parabolic. The horizontal speed of the ball is same as that of the car, therefore ball as well car travels equal horizontal distance, due to vertical speed, the ball follows parabolic path.

 

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Explanation – The boy throws the ball at an angle of 60.

Horizontal component of velocity 4cos $?$ = 10cos60

=10 (1/2)

=5m/s.

so horizontal speed of the car is same, hence relative velocity of car and ball in the horizontal direction will be zero.  

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Explanation – due to air resistance particle energy as well as horizontal component of velocity keep on decreasing making the fall steeper then rise. When we are neglecting air resistance path is parabola when we consider air resistance then path is asymmetric parabola.

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Explanation- when it be at position P, drops a bomb to hit a target T

Let $\theta$

Speed of the plane =720km/h = 720 $\times 5/18$ = 200m/s

Altitude of the plane P’T = 1.5km= 1500m

If bomb hits the target after time t then horizontal distance travelled by the bomb PP’=u $\times t$ =200t

Vertical distance travelled by the bomb P’T=1/2gt²

1500 = ½ (9.8)t²

So t²= 1500/4.9, t = $\sqrt[]{\frac{1500}{4.9}}=17.49s$

PP’=200 (17.49)m=

tan $\theta$ =P’T/P’P=1500/200 (17.49)=0.49287= tan23⁰12’

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Explanation – a) radius of earth =6400km= 6.4 $\times {10}^{6}m$

Time period = 1 day = 24 $\times 60\times 60$ = 86400s

Centripetal acceleration a= w²r= R(2 $\pi /T$ )²=4 $\pi$ ²R/T

= $\frac{4\times {\left(\frac{22}{7}\right)}^2\times 6.4\times {10}^6}{({24\times 60\times 60)}^2}$ = 0.034m/s²

b) time = 1yr=365 $\times 24\times 60\times 60$ days= 365=3.15 $\times {10}^{7}s$

centripetal acceleration = Rw²= $\frac{4{\pi }^{2}R}{T^2}$

^{$=\frac{4\times {\frac{22}{7}}^2\times 1.5\times {10}^{11}}{({3.15\times {10}^7)}^2}=5.97\times {10}^{-3}m/s$ 2}

$\frac{a_c}{g}=\frac{5.97\times {10}^{-3}}{9.8}=\frac{1}{1642}$

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Explanation – here A and B vectors are joint by head and tail. So C= A+B

(a) from fig iv it is clear that c=a+b

(a) From fig iii it is clear that c+b=a so a-c=b

(b) From fig I it is clear that b=a+c so b-a =c

(c) From ii it is clear that -c= a+b so a+b+c=0

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Explanation- given |A|=2 and |B|=4

a)|A $\times B$ |=AB sin $\theta$ = 0

so 2 $\times 4$ sin $\theta$ =0

so $\theta =0$ so it matches with iv

B) |A $\times$ B|= ABsin $\theta$ =8

2 $\times 4$ sin $\theta$ =8

So $\theta$ = 90 so it matches with option iii

c) |A $\times$ B|= ABsin $\theta$ =4

so $\theta$ =30 so it matches with option i

d) |A $\times$ B|= ABsin $\theta$ =4 $\sqrt[]{2}$

so $\theta$ =45 so it matches with option ii

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Explanation- - given |A|=2 and |B|=4

a) |A $\times B$ |=AB cos $\theta$ = 0

so 2 $\times 4$ cos $\theta$ =0

so $\theta =90$  so it matches with ii

b) |A $\times$ B|= ABcos $\theta$ =8

2 $\times 4$ cos $\theta$ =8

So $\theta$ = 0 so it matches with option i

c) |A $\times$ B|= ABcos $\theta$ =4

so $\theta$ =60 so it matches with option iv

d) |A $\times$ B|= ABcos $\theta$ =-8

so $\theta$ =180 so it matches with option iii

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Answer- b

Explanation-A = i+j

 B = i-j

 A.B=|A|B|cos $\theta$

(? +? ). (? -? ) = $\left(\sqrt[]{1+1}\right)\left(\sqrt[]{1+1}\right)$ cos $\theta$

Where $\theta$  is the angle between A and B

Cos $\theta$ = $\frac{1-0+0-1}{\sqrt[]{2\sqrt[]{2}}}$ = $\frac{0}{2}=0$

$\theta$  = 90^o

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Answer- d

Explanation- a scalar quantity is independent of direction hence has the same value for observers with different orientations of the axes.

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Answer- b 

Explanation – u= a? + b? as u in the first quadrant, hence both components a and b will be positive. For v= p? + q? , as it is positive x direction and located downward hence x component p will be positive and y component q will be negative.

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Answer- c

Explanation - we know that $\theta$ =15⁰ and R= 50m

Range, R= $\frac{u^{2}sin2\theta }{g}$

50 = $\frac{u^{2}sin2\times 15}{g}$

So u²= 980

So u = $\sqrt[]{980}=7\times 2\times \sqrt[]{5}$ m/s

So u= 14 $\times 2.23$ = 31.304m/s

$\theta$ = 45⁰, R= $\frac{u^{2}sin2\left(45\right)}{9.8}=\frac{u^2}{g}$

R= $\frac{({14\sqrt[]{5)}}^2}{g}$ = $\frac{14\times 14\times 5}{9.8}$ = $100m$

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Answer- b

Explanation – we know that impulse J= f. $?t$ = $?p$ , where F is force . $?t$  is time duration and $?p$  is change in momentum. As $?p$ is a vector quantity, hence impulse is also a vector quantity. Sometimes area can also be treated as vector.

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Answer- d

Explanation – speed = total distance travelled /time taken

Total distance travelled = path length

= speed $*$ time

We should be very careful with the fact, that speed is related with total distance covered not with displacement.

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Answer-b, d

Explanation- as given motion is two-dimensional motion and given that instantaneous speed v_o is positive constant. Acceleration is rate of change of velocity. hence it will also be in the plane of motion.

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Answer- b, d

Explanation – given A+B+C = 0

B $*\left(A+B+C\right)=B*0=0$

B $*A$ +B $*B+B*C$ =0

B $*A+0+B*C=0$

B $*A=-B*C$

A $*B=B*C$

(A $*B$ ) $*C=\left(B*C\right)*C$

It cannot be zero

(b) (A $*B$ ).C= (B $*AC$ ).C=0 . if b|C then B $*C$ =0 then (B $*C$ ) $*C=0$

(c) (A $*B$ )=X=ABsin $?X$ . The direction of X is perpendicular to the plane containing A and B (A $*B$ ) $*C=X*C.$

(d) if c²= A²+B², then angle between A and B is 90⁰

(A $*B$ ).C= (AB sin90⁰X).C=AB (X.C)

= ABC cos90⁰= 0

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Answer- a, b

Explanation - |A+B|= |A|or |A+B|²=|A|²

|A|² +|B|²+2|A|B|cos $\theta$ = |A|²

|B| (|B|+2|A|cos $\theta$ )= 0

|B|=0 or |B|+2|A|cos $\theta$ =0

Cos $\theta$ = $\frac{-\left|B\right|}{2\left|A\right|}$

If A and B are antiparallel then $\theta$ =180

-1= $\frac{\left|B\right|}{2\left|A\right|}=\left|B\right|=2\left|A\right|$

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Answer- a,b,c

Explanation – H= $\frac{u^2{sin}^2\theta }{2g}$

H₁=V_o²sin^{2 $\theta$} ₁/2g  , H₂=V_o²sin^{2 $\theta$} ₂/2g

H₁>H₂

V_o²sin^{2 $\theta$} ₁/2g= V_o²sin^{2 $\theta$} ₂/2g

Sin^{2 $\theta$} ₁>sin^{2 $\theta$} ₂

Sin^{2 $\theta$} ₁ – sin² $\theta$ ₂>0

(Sin $\theta$ ₁ – sin $\theta$ ₂)( Sin $\theta$ ₁ + sin $\theta$ ₂)>0

Sin $\theta$ ₁>sin $\theta$ ₂ or ₁ >₂

T= $\frac{2usin\theta }{g}=\frac{2v_{o}sin\theta }{g}$

T₁= $\frac{2v_{o}sin{\vartheta }_1}{g}$   , T₂= $\frac{2v_{o}sin{\vartheta }_2}{g}$

T₁> T₂

R= $\frac{u^{2}sin2\theta }{g}=\frac{v_o^{2}sin2\theta }{g}$

Sin $\theta$ ₁>sin $\theta$ ₂

Sin2 $\theta$ ₁> sin2 $\theta$ ₂

$\frac{R_1}{R_2}=\frac{\mathrm{S}\mathrm{i}\mathrm{n}2{\theta }_1}{sin2{\theta }_2}1$

R₁>R₂

Total energy for the first particle

U₁=K.E+P.E=1/2m_{1 $v_o^2$}

U₂= K.E+P.E= 1/2m_{2 $v_o^2$}

Total energy for the second particle

So m₁= m₂ then U₁=U₂

So m₁>m₂ then U₁>U₂

So m₁2 then U1

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Answer- c

Explanation– as the given track y=x² is a frictionless track thus total energy will be same throughout the journey.

Hence total energy at A = total energy at P . at B the particle is having only Ke but at P some KE is converted to P

Hence (KE)_B = (KE)_P

Total energy at A = PE= total energy at B = KE= total energy at P

= PE+KE

Potential energy at A is converted to KE and PE at P hence

(PE)P< (PE)A

Hence (height)P= (height)A

As height of p < height of A

Hence path length AB > path length BP

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Answer- a, c 

Explanation – as we know average acceleration is a_av= $\frac{?v}{?t}=\frac{v_2-v_1}{t_2-t_1}$

But when acceleration is not uniform V_av is not equal to v₁+v₂/2

So we can write $?v=\frac{?r}{?t}$

$?r=r_2-r_1$ = v₂-v₁ (t₂-t₁)

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Answer- a, b, c

Explanation- (i) Speed will constant throughout

(ii) Velocity will be tangential in the direction of motion

(iii) Centripetal acceleration will be a= v²/r, will always be towards centre of the circular path.

(iv) Angular momentum is constant in magnitude and direction out of the plane perpendicularly as well.

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Explanation- |A+B|=|A-B|

$\sqrt[]{\left|{A|}^2+\left|{B|}^2+2\right|A\right|\left|B\right|cos\theta }=\sqrt[]{\left|{A|}^2+\left|{B|}^2-2\right|A\right|\left|B\right|cos\theta }$

$\left|{A|}^2+\left|{B|}^2+2\right|A\right|\left|B\right|cos\theta$  = $\left|{A|}^2+\left|{B|}^2-2\right|A\right|\left|B\right|cos\theta$

4|A|B|cos $\theta$ =0

|A|²+|B|²cos $\theta$ =0

A=0 or B=0 so $\theta =90$ . so A perpendicular B

According to question, we can write

10 = $\frac{1}{2}a{t}^2...............\left(1\right)and$

$10+x=\frac{1}{2}a{\left(2t\right)}^2................\left(2\right)$

$\Rightarrow X=\frac{1}{2}A\left[{\left(2t\right)}^2-t^2\right]=3\left(\frac{1}{2}a{t}^2\right)=30m$

According to question, we can write

 $d=d_0\left(1+\alpha \Delta T\right)\Rightarrow 6.241=6.230\left(1+1.4\times 10^{-5}\times \Delta T\right)$

$\Rightarrow T=126.18+27=153.18°C$

According to question, we can write

$F_{CA}=F_{CB}=\frac{KqQ}{x^2+{\left(\frac{d}{2}\right)}^2}\Rightarrow F=2F_{CA}cos\theta =\frac{2KqQx}{{\left[x^2+{\left(\frac{d}{2}\right)}^2\right]}^{\frac{3}{2}}}$

For maxima of force

$\frac{dF}{dx}=0,so$

$x=\frac{d}{2\sqrt[]{2}}$

$\frac{sin{\theta }_C}{V_B}=\frac{sin90°}{V_A}\Rightarrow sin{\theta }_C=\frac{V_B}{V_A}=\frac{1.5\times 10^{10}}{2.0\times 10^{10}}=\frac{3}{4}$

According to question, we can write

$\theta >\theta =si{n}^{-1}\left(\frac{3}{4}\right)$

Please find the solution below:

According to question, we can write

10 = $\frac{1}{2}a{t}^2...............\left(1\right)and$

$10+x=\frac{1}{2}a{\left(2t\right)}^2................\left(2\right)$

$\Rightarrow X=\frac{1}{2}A\left[{\left(2t\right)}^2-t^2\right]=3\left(\frac{1}{2}a{t}^2\right)=30m$

According to question, we can write

 $d=d_0\left(1+\alpha \Delta T\right)\Rightarrow 6.241=6.230\left(1+1.4\times 10^{-5}\times \Delta T\right)$

$\Rightarrow T=126.18+27=153.18°C$

According to question, we can write

$F_{CA}=F_{CB}=\frac{KqQ}{x^2+{\left(\frac{d}{2}\right)}^2}\Rightarrow F=2F_{CA}cos\theta =\frac{2KqQx}{{\left[x^2+{\left(\frac{d}{2}\right)}^2\right]}^{\frac{3}{2}}}$

For maxima of force

$\frac{dF}{dx}=0,so$

$x=\frac{d}{2\sqrt[]{2}}$

$\frac{sin{\theta }_C}{V_B}=\frac{sin90°}{V_A}\Rightarrow sin{\theta }_C=\frac{V_B}{V_A}=\frac{1.5\times 10^{10}}{2.0\times 10^{10}}=\frac{3}{4}$

According to question, we can write

$\theta >\theta =si{n}^{-1}\left(\frac{3}{4}\right)$

Please find the solution below:

The light wave contains two lights of different frequencies, so

$E_1=h{\upsilon }_1=\frac{4.14\times 10^{-15}\times 6\times 10^{15}}{2\pi }=3.96eV,and$

$E_2=h{\upsilon }_2=\frac{4.14\times 10^{-15}\times 9\times 10^{15}}{2\pi }=5.92eV$

Maximum kinetic energy of the photoelectron = 5.9-2.5 = 3.42 eV

According to Law of discharging of capacitor, we can write

$Q=Q_0{e}^{-\frac{t}{\tau }}\Rightarrow \frac{Q_2}{8}=Q_0{e}^{-\frac{t_2}{\tau }}\Rightarrow t_{2}3\tau \mathcal{l}n\left(2\right),and$

$U=\frac{Q^2}{2C}=\frac{Q_0^2}{2C}{e}^{-\frac{2t}{\tau }}\Rightarrow \frac{1}{2}\left(\frac{Q_0^2}{2C}\right)=\frac{Q_0^2}{2C}{e}^{-\frac{2t}{\tau }}\Rightarrow t_1=\frac{\tau }{2}\mathcal{l}n\left(2\right)$

$\Rightarrow \frac{t_1}{t_2}=\frac{1}{6}$

Process-AB $\Rightarrow$ Isobaric,

Process-AC $\Rightarrow$ Isothermal, and

Process-AD? Adiabatic

W₂ < W₁ < W₃

If currents are flowing in same direction, magnetic field will cancel each other, so the currents must flowing in opposite direction

$B_P=\frac{{\mu }_{0}I}{2\pi r}\times 2$

$\Rightarrow 300\times 10^{-6}=\frac{4\pi \times 10^{-7}}{2\pi \times 4\times 10^{-2}}\times 2$

$\Rightarrow$ I = 30 A

According to question, we can write

$d=\sqrt[]{2hR}\Rightarrow \frac{d_2}{d_1}=\sqrt[]{\frac{h_2}{h_1}}\Rightarrow h_2={\left(\frac{d_2}{d_1}\right)}^2{h}_1={\left(\frac{2}{1}\right)}^2\times 125=500m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

According to question, we can write

$d=\sqrt[]{2hR}\Rightarrow \frac{d_2}{d_1}=\sqrt[]{\frac{h_2}{h_1}}\Rightarrow h_2={\left(\frac{d_2}{d_1}\right)}^2{h}_1={\left(\frac{2}{1}\right)}^2\times 125=500m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

According to question, we can write

$\omega =\pi =\sqrt[]{\frac{g}{\mathcal{l}}}\Rightarrow \mathcal{l}=\frac{g}{{\pi }^2}=\frac{9.8}{{\left(3.14\right)}^2}=0.99395=99.4cm$

According to question, we can write

Total moles of gas = n = n_Oxygen + n_Oxygen = $\frac{16}{2}+\frac{128}{32}=12moles$

Volume of gas = 12 × 22.4 litre = 268.8 litre = 2.688 × 10⁵ cm3 $\approx 27\times 10^{4}c{m}^3$

According to Lorentz’s Force, we can write

$\overrightarrow{F}={\overrightarrow{F}}_{Electric}+{\overrightarrow{F}}_{Magnetic}=q\overrightarrow{E}+q\left(\overrightarrow{v}\times \overrightarrow{B}\right),so$

Statement I is correct but Statement II is incorrect

According to Newton’s laws of motion, we can write

4a = 4g – T . (1), and

40a = T - fk = T $\mu$  (40g) ……………… (2)

Adding equations (1) and (2), we can write

44a = 40 – 0.02 × (400) = 32

$\Rightarrow a=\frac{32}{44}=\frac{8}{11}m/s^2$

According to conservation of energy, we can write

Gain in kinetic energy = Loss in potential energy

K_f – K_in = U_in - U_f

K - = mgy – mg (y – y₀) = mgy₀

According to Newton’s law of motion, we can write

ma = mg – N = mg $\frac{mg}{4}=\frac{3mg}{4}$

$\Rightarrow a=\frac{3g}{4}$

According to relation between field and potential, we can write

$\overrightarrow{E}=-\frac{dV}{dx}\left(\widehat{i}\right)=-\frac{d\left(3x^2\right)}{dx}\left(\widehat{i}\right)=-6x\left(\widehat{i}\right)$

${\overrightarrow{E}}_{\left(1,0,3\right)}=-6\left(\widehat{i}\right)N/C$