Q

Interference pattern is observed at ‘P’ due to superimposition of two rays coming out
from a source ‘S’ as shown in the figure.The value of ‘’ for which maxima is
obtained at ‘P’ is : (R is perfect reflecting surface) :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

V Vakul Arora

As we have learned

Lloyd's mirror Experiment -

$S_{2}P-S_{1}P= n\lambda$

$S_{2}P-S_{1}P= \left ( n+1/2 \right )\lambda$

- wherein

minima

maxima

For maxima at P

$\Delta x= (n-1/2 )\lambda$

$x \cos 30 \degree = l \\ \Rightarrow x = \frac{2l }{\sqrt 3 }$

$\Delta x = \frac{2l}{\sqrt 3}+ \frac{2l}{\sqrt 3 } - 2l = (n- 1/2 )\lambda$

Or

$\frac{4l}{\sqrt3}-2l = (n-1/2 )\lambda$

or

$l = \frac{(2n-1)\lambda \sqrt 3 }{4(2-\sqrt 3 )}$

Option 1)

Option 2)

Option 3)

Option 4)

295 Views

The circuit has two oppositely connect ideal diodes in parallel. What is the current following in the circuit?

• Option 1)

1.33 A

• Option 2)

1.71 A

• Option 3)

2.00 A

• Option 4)

2.31 A.

A Avinash

As we have learnt,

P -N junction as diode -

It is a one way device. It offers a low resistance when forward biased and high resistance when reverse biased.

- wherein

R = 0, Forward

$\rightarrow \infty$ Reverse

In the figure D, is reverse biased, hence no current will flow through D.

$\Rightarrow I = \frac{V}{R} = \frac{12V}{6\Omega} = 2 A$

Option 1)

1.33 A

Option 2)

1.71 A

Option 3)

2.00 A

Option 4)

2.31 A.

313 Views

In a full wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be

• Option 1)

100 Hz

• Option 2)

70.7 Hz

• Option 3)

50 Hz

• Option 4)

25 Hz

A Avinash

As we have learnt,

Full wave rectifier -

It is type of rectification in which output is obtained for both the halves of a.c.

- wherein

Frequency of output is double to that of input

In full wave rectifier

Input                                                                                   Output

Time period becomes half

Frequency gets doubled

Frequency is 100 Hz

Option 1)

100 Hz

Option 2)

70.7 Hz

Option 3)

50 Hz

Option 4)

25 Hz

271 Views

If in a $p-n$  junction diode, a square input signal of 10 V is applied as shown

Then the output signal across $R_L$ will be

• Option 1)

• Option 2)

• Option 3)

• Option 4)

A Avinash

As we have learnt,

P -N junction as diode -

It is a one way device. It offers a low resistance when forward biased and high resistance when reverse biased.

- wherein

R = 0, Forward

$\rightarrow \infty$ Reverse

When input is -5V, then diode is reverse biased and hence output is zero

When input is +5V, then output is also +5V

Option 1)

Option 2)

Option 3)

Option 4)

242 Views

The machine as shown has 2 rods of length 1 m connected by a pivot at the top.  The end of one rod is connected to the floor by a stationary pivot and the end of the other rod has a roller that rolls along the floor in a slot.  As the roller goes back and forth, a 2 kg weight moves up and down.  If the roller is moving towards right at a constant speed, the weight moves up with a :

• Option 1)

constant speed

• Option 2)

decreasing speed

• Option 3)

increasing speed

• Option 4)

speed which is      th of that of the roller when the weight is 0.4 m above the ground

S subam

From $\Delta ABC \\ y = \sqrt{1- (\frac{x}{2})^2}$

$\frac{dy}{dt}= \frac{d}{dt}(\frac{\sqrt{4x^2}}{2}) = 1/2 \frac{d}{dx } (\sqrt { 4x^2})dx /dt$

$= \frac{1}{2}\cdot \frac{1}{2\sqrt{4- x^2}}(-2x )dx/dt$

$\frac{dy}{dt} = \frac{-x }{2 \sqrt {4 x^2 }}\cdot dx/dt$

dx /dt = constant

$\Rightarrow \frac{dy}{dt} = \frac{-x }{2 \sqrt {4 x^2 }}\cdot dx/dt$

$\Rightarrow \frac{dy}{dt} = \frac{-x }{2 \sqrt {\frac{4}{x^2}-1 }}\cdot dx/dt$

As with motion of roller , x decreases

$\therefore \sqrt{\frac{4}{x^2}-1 }\: \: increases$

dy/dt decreases

weight will move with decreasing speed

Option 1)

constant speed

Option 2)

decreasing speed

Option 3)

increasing speed

Option 4)

speed which is      th of that of the roller when the weight is 0.4 m above the ground

519 Views

A line drawn through the point P(4, 7) cuts the circle x2+y2=9 at the points A and B. Then PA⋅PB is equal to :

• Option 1)

53

• Option 2)

56

• Option 3)

74

• Option 4)

65

S satyajeet

As we learnt

Length of a tangent -

$L=\sqrt{x{_{1}}^{2}+y{_{1}}^{2}+2gx_{1}+2fy_{1}+c}$

- wherein

Length of tangent from a external point $(x_{1},y_{1})$  to circle $x^{2}+y^{2}+2gx+2fy+c=0$

Since we know,

$PA.PB=PT^2$

$PT=\sqrt{4^2+7^2-9}$

$\Rightarrow PT^2= \left ( \sqrt{56} \right )^{2}= 56$

Option 1)

53

Option 2)

56

Option 3)

74

Option 4)

65

185 Views

A spring of unstretched length l has a mass m with one end fixed to a rigid support. Assuming spring to be made of a uniform wire, the kinetic energy possessed by it if its free end is pulled with uniform velocity is :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

S SudhirSol

As we have learned

Kinetic energy -

$k= \frac{1}{2}mv^{2}$

- wherein

$m\rightarrow mass$

$v\rightarrow velocity$

Kinetic Energy is always positive.

fixed end is at rest and free end is moving with speed v

Velocity at distance x from free end $v = \frac{V}{L}x$

Kinetic energy of mass dm at x = x is K.E = $1/2 (dm )\left [ \left ( \frac{V}{L} \right )^2 \right ]^L$

$\frac{1}{2}\left ( \frac{M}{L}dx \right )\frac{V^2}{L^2}x^2$

$\frac{M}{2L^3}v^2x^2 dx$

Total kinetic energy = $\int_{0}^{L}\frac{Mv^2}{2L^3}x^2dx = \frac{Mv^2}{2L^3}\cdot \frac{L^3}{3}= \frac{1}{6}Mv^2$

Option 1)

Option 2)

Option 3)

Option 4)

146 Views

Two soap bubbles coalesce to form a single bubble. If V is the subsequent change in volume of contained air and S the change  in total surface area, T is the surface  tension and P atmospheric pressure, which  of the following relation is correct ?

• Option 1)

• Option 2)

• Option 3)

• Option 4)

S SudhirSol

As we have learned

Change in Pressure of bubble in air -

$\Delta P=\left ( \frac{2T}{R} \right )\times 2= \frac{4T}{R}$

- wherein

T- Temperature

$\left ( P + \frac{4 T}{r_1} \right )\left ( \frac{4 \pi }{3}r_{1}^{3} \right )+ \left ( P + \frac{4T}{r_2} \right )\frac{4 \pi }{3}r_{2}^{3}= \left ( P + \frac{4T}{r} \right )\frac{4\pi}{3}r^3$

$PV+ \frac{4T}{3}(4 \pi r_{1}^{2}+4 \pi r_{2}^{2}-4 \pi r^2)= 0$

$\Rightarrow 3PV + 4TS = 0$

Option 1)

Option 2)

Option 3)

Option 4)

785 Views

A tank with a small hole at the bottom has been filled with water and kerosene (specific gravity 0.8). The height of water is 3 m and that of kerosene 2 m. When the hole is opened the velocity of fluid coming out from it is nearly : (take g=10 ms-2 and density of water =103 kg m-3)

• Option 1)

10.7 ms-1

• Option 2)

9.6 ms-1

• Option 3)

8.5 ms-1

• Option 4)

7.6 ms-1

S SudhirSol

As we have learned

Torricelli's Theorem / Velocity of Efflux -

In fluid dynamics relating the speed of fluid flowing out of an orifice.

- wherein

According to torcellli theorem

velocity of efflux = $\sqrt{2gh }$

Left density of water is $\rho w$

Density of kerosene = 0. 8 $\rho w$

Apply bernoulli between A and B

$\rho _a + \rho _wg\times 3+(0.8\rho _w)g\times 2=\rho _a+0 +1/2 \cdot \rho _wv^2$

$v^2 = 9.2 g \Rightarrow v = \sqrt{92 }m/s = 9.6 m/s$

Option 1)

10.7 ms-1

Option 2)

9.6 ms-1

Option 3)

8.5 ms-1

Option 4)

7.6 ms-1

102 Views

A cylindrical vessel of cross-section A contains water to a height h.There is a hole in the bottom of radius 'a'.The time in which it will be emptied is :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

S SudhirSol

As we have learned

Torricelli's Theorem / Velocity of Efflux -

In fluid dynamics relating the speed of fluid flowing out of an orifice.

- wherein

Velocity of eflux of water = $\sqrt { 2gh }$

$\frac{dV}{dt}=-A\frac{dh}{dt}= \pi a^2 \sqrt { 2gh }$

$\int_{h}^{0}\frac{dh}{h}= - \pi \frac{a^2\sqrt {2g}}{A} \int_{0}^{t_0}dt$

$t = \frac{A}{\pi a^2}\cdot \sqrt{\frac{2h}{g}}$

Option 1)

Option 2)

Option 3)

Option 4)

170 Views

The figure shows a system of two concentric spheres of radii $r_{1}\: and\: r_{2}$ and kept at temperatures $T_{1}\: and\: T_{2}$ respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to

• Option 1)

$\frac{r_{1}r_{2}}{\left ( r_{2}-r_{1} \right )}$

• Option 2)

$\left ( r_{2}-r_{1} \right )$

• Option 3)

$\frac{\left ( r_{2}-r_{1} \right )}{r_{1}r_{2}}$

• Option 4)

$\ln \left ( \frac{r_{2}}{r_{1}} \right )$

S SudhirSol

As we have learned

Thermal Conductivity -

$Q=\frac{KA(\theta_{1}-\theta_{2})t}{l}$

K = thermal conductivity

- wherein

Rate of flow of heat $\frac{d\theta }{dt} = \frac{KA\Delta T}{L}$

We have a spherical shell of radius r and thickness dx

$A = 4 \pi r^2 , L =dr \\ \frac{d\theta }{dt}= \frac{K4 \pi r^2 dt }{dr }$

$4 \pi K \int_{T_1}^{T_2}dT= \frac{d\theta }{dt} \int_{r_1}^{r_2}\frac{dr}{r^2} = \frac{d\theta }{dt} \left ( \frac{1}{r_1}-\frac{1}{r_2} \right )$

$4 \pi K(T_2-T_1)= \frac{d\theta }{dt} \left ( \frac{r_2r_1}{r_2-r_1} \right )$

$\therefore \frac{d\theta }{dt} \propto \frac{r_2r_1}{r_2-r_1}$

Option 1)

$\frac{r_{1}r_{2}}{\left ( r_{2}-r_{1} \right )}$

Option 2)

$\left ( r_{2}-r_{1} \right )$

Option 3)

$\frac{\left ( r_{2}-r_{1} \right )}{r_{1}r_{2}}$

Option 4)

$\ln \left ( \frac{r_{2}}{r_{1}} \right )$

99 Views

Water is flowing at a speed of 1.5 ms-1 through a horizontal tube of cross-sectional area 10-2 m2 and you are trying to stop the flow by your palm. Assuming that the water stops immediately after hitting the palm, the minimum force that you must exert should be (density of water=103 kgm-3).

• Option 1)

15 N

• Option 2)

22.5 N

• Option 3)

33.7

• Option 4)

45 N

S SudhirSol

As we have learned

Equation of Continuity -

Mass of the liquid entering per second at A = mass of the liquid leaving per second at B.

a1 v1 = a2 v2

- wherein

a1  and abe the area of cross section.

Let us say speed of water is v and area of cross section is A

In one second mass of water that flows is $m = \rho A v$

F = momentum change per second

$mv = \rho A v ^2= 10 ^3 \times 10^{-2}\times (1.5 )^2= 22.5 N$

Option 1)

15 N

Option 2)

22.5 N

Option 3)

33.7

Option 4)

45 N

134 Views

An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be length of the air column above mercury in the tube now ?

(Atmospheric pressure =76 cm of Hg)

• Option 1)

16 cm

• Option 2)

22 cm

• Option 3)

38 cm

• Option 4)

6 cm

S SudhirSol

As We have learned

Absolute Pressure -

$P= P_{0}+\rho\: gh$

- wherein

$P\rightarrow hydrostatics \: Pressure$

$P_{0}\rightarrow atmospheric \: Pressure$

For air trapped in tube T = constant

$\Rightarrow P_1 V_1 = P_2V_2 \Rightarrow P_1 = P_{atm}= \rho g(76) \\V_1 = A \pi 8 cm$

$P_2 = P_a - \rho g(54- x )= \rho g (76-54+x )\\P_2 = \rho g (22+x )\\V_2 = A \cdot x \\\Rightarrow \rho g(76 )\times (8A )= \rho g(22+x )(Ax )\\\Rightarrow x^2 +22x - 76 \times 8 = 0 \\\Rightarrow x = 16 cm$

Option 1)

16 cm

Option 2)

22 cm

Option 3)

38 cm

Option 4)

6 cm

147 Views

A thin bar of length L has a mass per unit  length, that increases linearly with distance from one end.If its total mass is M and its mass per unit length at the lighter end is , then the distance of the centre of mass from the lighter end is :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

S SudhirSol

As we have learned

Centre of Mass of a continuous Distribution -

$x_{cm}=\frac {\int xdm}{\int dm}, \; \;y_{cm}=\frac{\int ydm}{\int dm}, \;z_{cm}=\frac{\int zdm}{\int dm}$

- wherein

dm is mass of small element. x, y, z are coordinates of dm part.

$\lambda = \lambda _0 +kx$

k is some constant

Total mass = $\int dm = \int_{0}^{L}\lambda dx$

$M = \int_{0}^{L}(\lambda_0 +k x )dx = \lambda _0 L + \frac{k L^2}{2}.......(1)$

$x_c_m = \frac{\int xdm}{\int dm } = \frac{\int_{0}^{L}x (\lambda dx)}{M}$

$= \frac{\int_{0}^{L}(\lambda _0+kx )\cdot xdx }{M}= \frac{\lambda _0\cdot \frac{L^2}{2}+ k \frac{L^3}{3}}{M}$

$x_c_m = \frac{L^2\left [ \frac{\lambda _0}{2}+\frac{KL}{3} \right ]}{M}$

From equation (1) $KL = \left ( \frac{M}{L}-\lambda _0 \right )\cdot 2$

$x_c_m = \frac{L^2}{M} \left [ \frac{\lambda _0}{2}+\frac{2M}{3L} - \frac{2 \lambda_0}{3}\right ]$

$= \frac{L^2}{M} \left [ \frac{3\lambda _0L+2M-4\lambda _0L}{6L}\right ]$

$= \frac{L}{6M}\cdot (2M - \lambda _0L)= L/3 - \frac{\lambda _0L^2}{6M}$

Option 1)

Option 2)

Option 3)

Option 4)

92 Views

A hoop of radius r and mass m rotating with an angular velocity   is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velociy of the centre of the hoop when it ceases to slip ?

• Option 1)

• Option 2)

• Option 3)

• Option 4)

S SudhirSol

As we have learned

Law of conservation of angular moment -

$\vec{\tau }= \frac{\vec{dL}}{{dt}}$

- wherein

If net torque is zero

i.e. $\frac{\vec{dL}}{{dt}}= 0$

$\vec{L}= constant$

angular momentum is conserved only when external torque is zero .

When loop ceases to slip $V_{em}= rw ......(1)$

angular momentum is conserved about point of contact P

$L_i = L_f$

$L_i =$ L of c.m + L about C.M

= 0 + Iw = $Mr ^ 2 w_0$

$L_f =$  L of cm + L about C.M

= Mvr + $(Mr^2)w = 2 Mvr (\because V_{cm}=rw )$

$\Rightarrow 2 M V_{cm} r = M r^2 w_0$

$\Rightarrow v_{cm }= \frac{w_0 r }{2}$

Option 1)

Option 2)

Option 3)

Option 4)

74 Views
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