InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 3101. |
Which of the following represents electric polarisation ? |
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Answer» `P=(1)/(sigma_(P))` |
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| 3102. |
When a force 1 N acts on 1kg mass at rest for 1s, its final momentum is P. When 1N force acts on 1kg mass at rest through a distance Im, its final momentum is P. The ratio of P to P is |
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Answer» `1 : 1` |
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| 3103. |
A particle is projected with initial velocity u = 10 m//s from a point ( 10m, 0m, 0m ) m along z-axis. At the same time , another particle was also projected with velocity v ( hat(i) - hat(j) + hat( k)) m//s from ( 0m, 10m, 0m). It is given that they collide at some point in the space . Then find the value of v. [ vec(g) = 10 m//s^(2) ( - hat(k))] |
| Answer» Answer :A | |
| 3104. |
In n-p-n transistor circuit, the collector current is 20 mA. If 90% of the electrons emitted reache the collector, then the |
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Answer» Emitter currect will be about 16 mA |
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| 3105. |
When S, close bulb glows Instantaneously when S_2 closes there is a delay in glowing the bulb. Explain the phenomenon self induction regarding above experiment |
| Answer» Solution :When a.c. current is PASSED through the coil, a change in FLUX is produced. This change in flux PRODUCES a BACK e.m.f. in the coil. The phenom ena of production of back emf is called SELF induction. | |
| 3106. |
For a particle executing SHM, which of the following statements does not hold good ? |
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Answer» the total energy of the PARTICLE always remains the same |
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| 3107. |
The mean path length of alpha-particles in air under standard conditions is defined by the formula R=0.98.10^(27)v_(0)^(3) cm, this formula, find for an alpha-particle with initial kinetic enrgy 7.0 MeV: (a) its mean path length, (b) the average number of ion pairs formed by the given alpha-particle over the whole path R as well as over its first half, assuning the ion pair formation energy to be equql to 34 eV. |
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Answer» Solution :(a) For an ALPHA particle with initial `K.E 7.0 MeV`, the initial VELOCITY is `V_(0)=sqrt((2T)/(M alpha))` `=sqrt(2xx7xx1.602xx10^(-6))/(4xx1.672xx10^(-24))` `= 1.83xx10^(9)cm//sec` Thus `R=6.02 cm` (b) Over the whole path the number of ion PAIRS is `(7xx10^(6))/(34)= 2.06xx10^(5)` Over the first half of the path:- We write the formula for the mean path as `R alphaE^(3//2)` where `E` is the initial energy. Thus if the energy of the `alpha` particle after traversing the first half of the path is `E_(1)` then `R_(0)E_(1)^(3//2)=(1)/(2)R_(0)E_(0)^(3//2) or E_(1)= 2^(-2//3)E_(0)` Hence number of ion pairs formed in the first half of the path LENGTH is `(E_(0)-E_(1))/(34eV)=(1-2^(-2//3))xx2.06xx10^(5)= 0.76xx10^(5)` |
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| 3108. |
A simple pendulum of period T has a metal bob which is negatively charged. If it is allowed to oscillate above a positively charged metal plate, its period will be |
| Answer» Answer :B | |
| 3109. |
Some rich people also buy ……………………..furniture. |
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Answer» MAHOGANY furniture |
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| 3110. |
A 10 kg collar is attached to a spring (spring constant 600 N/m.), it slides without friction over a horizontal rod. The collar iis displaced from the equilibrium position by 20 cm and released. What is the speed of the oscillation ? |
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Answer» `sqrt(60)xx0.2" m"//"s"` `:. (1)/(2)KX^(2)=(1)/(2)mv^(2)` `:.""v=sqrt((kx^(2))/(M))` `v=sqrt((600)/(10)).(20)/(100)` `v=sqrt(60)xx0.2" m"//"s"` So correct CHOICE is (a). |
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| 3111. |
Statement-I : The velocity of sound in a medium increases with the presence of moisture. Statement -II : Velocity of sound does not depend upon the nature of medium . |
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Answer» Statement -I is true, Statement -II is true and |
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| 3112. |
Derive an expression for the escape speed of a body from the surface of the Earth. Express it in terms of the mean density of the Earth. |
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Answer» Solution :For projection speed EQUAL to the escape speed `(v_(E))`, the kinetic energy of the body equals the binding energy. `therefore (1)/(2) mv_(e)^(2) = (GMM)/(R )` where m is the mass of the body, G is the gravitational CONSTANT, M is the mass of the Earth and R is the radius of the Earth. `therefore v_(e)=sqrt((2GM)/(R ))` The mean density of the Earth, `rho = ("mass")/("volume") = (M)/((4)/(3)piR^(3))` `therefore M = (4)/(3) pi R^(3) rho` `thereforev_(e)=sqrt((2G)/(R)xx(4)/(3)piR^(3)rho)=2Rsqrt((2piGrho)/(3))` |
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| 3113. |
In the figure, each segment (side of small triangle) has resistance R and the wire used in the circumference of the circle has negligible resistance. find equivalent resistance between point O and A. |
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Answer» |
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| 3114. |
A proton has a mass 1.67 xx 10^(-27) kg and charge + 1.6 xx 10^(-19)C. If the proton is accelerated through a potential difference of million volts, then the kinetic energy is |
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Answer» `1.6 XX 10^(-15)` J |
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| 3115. |
A uniform rope having some mass hangs vertically from a rigid support. A transverse wave pulse is produced at the lower end. The speed (v) Of the wave pulse varies with height (h) from the lower end as: |
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Answer»
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| 3116. |
What is the advantage of a meter bridge over a Wheatstone bridge? |
| Answer» SOLUTION :Meter BRIDGE is an IMPROVEMENT on Wheatstone.s bridge in the sense that we can change RESISTANCE continuously. | |
| 3117. |
A resistor of 100 Omega and a capacitor of 250 muF are connected in series to a 220 V , 50 Hz ac source. Calculate the current in the circuit |
| Answer» SOLUTION :`Z APPROX 162.0 OMEGA`,I=1.358 A | |
| 3118. |
Draw the circuit diagram of an illuminated photodiode in reverse bias. How is photodiode used to measure light intensity ? |
Answer» Solution :Circuit DIAGRAM of an ILLUMINATED photodiode :  Explanation : The magnitude of the photocurrent depends on the INTENSITY of incident light (photocurrent is PROPORTIONAL to incident light intensity). THUS photodiode can be used to measure light intensity |
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| 3119. |
A signal of 5 KHz frequency is amplitude modulated of the resultant signal is /are: |
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Answer» 2MHZ als |
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| 3120. |
A spherical coductor of radius 12cm has a charge of 1.6 xx 10^(-7)C distributed uniformly on its surface. What is the electric field (a) inside (b) just outside the sphere (c), ata point 18cm from the centre of the sphere ? |
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Answer» SOLUTION :`R=12 XX 10^(-2)m, Q=1.6 xx 10^(-7)C` a. Zero B. `E=(9 xx 10^(9) xx 1.6 xx 10^(-7))/((12 xx 10^(-2))^(2))=(9 xx 10^(9) xx 1.6 xx 10^(-7))/(12 xx 12 xx 10^(-4))=10^(5)NC^(-1)` c. Here `R=18 xx 10^(-2)m` Hence `E=(9 xx 10^(9)xx 1.6 xx 10^(-7))/(18 xx 18 xx 10^(-4))=4/9 xx 10^(5) approx 4.4 xx 10^(-4)NC^(-1)` |
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| 3121. |
A pendulum of mass m hangs from a support fixed to a trolley. The direction of the string (i.e., angle theta) when the trolley rolls up a plane of inclination a with acceleration 'a' is |
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Answer» ZERO |
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| 3122. |
If an oil drop is poured on water, then it will |
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Answer» REMAIN SPHERICAL on water surface |
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| 3123. |
A body is projected up along an inclined plane from the bottom with speed V_(1).If it reachesthe bottom of the plane with a velocity V_(2),find (v_(1)//v_(2)) if thetais the angle of inclination with the horizontal and mu be the coefficient of. |
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Answer» `(SIN theta + MU COS theta)/(sin theta - mu cos theta)` |
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| 3124. |
The current through a coil self inductance L = 2 mH is given by I=t^(2)e^(-1) at time t. How long it will take to make the e.m.f. zero ? |
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Answer» Solution :`I=t^(2)e^(-t)` `therefore (dl)/(dt)=2te^(-t)-t^(2)e^(-t)=te^(-t)(2-t)` The induced emf is `epsilon = -L(dI)/(dt)` ACCORDING to GIVEN problem, `epsilon = 0` `rArr (dI)/(dt)=0 ""` (Since `L ne 0`) or `e^(-t)t(2-t)=0`.Either t = 0 or t = 2 s t = 2 matches with the option (b). |
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| 3125. |
If the Young's experiment is conducted in a medium of refractive index mu, the fringewidth is given by ___. |
| Answer» SOLUTION :`BETA = (D LAMBDA)/(MU d) ` | |
| 3126. |
If f_(0) and f_(e) be the focal lengths of the objective and eyepiece respectively of an astronomical telescope, its magnifying power will be |
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Answer» `f_(0)+f_(E)` |
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| 3127. |
A resistor of 100 Omega and a capacitor of 250 muF are connected in series to a 220 V , 50 Hz ac source. Calculate the voltage (rms) across the resistor and the capacitor . Is the algebraic sum of these voltages more than the sourcevoltage ? If yes, resolve the paradox. |
| Answer» SOLUTION :`V_R`= 135.8 V , `V_C`= 184.4 V , `V_(R+C)`= 229 V | |
| 3128. |
When a ceiling fan is switched on,it makes 10 rotation in the first 3 seconds. How many rotation will it make in the next 3 seconds?(Assume uniform angular acceleration) |
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Answer» 10 |
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| 3129. |
If no element of set A is connected to any ement of Set B then relation is called |
| Answer» Answer :B | |
| 3130. |
Draw a neat labeled diagram to obtain an inverted image in the case of a total reflecting prism. |
Answer» SOLUTION :
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| 3131. |
In the situation described in Problem 63, find the condition that a charge kept on x = 2m may be in a stable equilibrium. If the charge is of magnitude 1 mu C and has a mass of 3 g, find the time period of small oscillations of the body from its equilibrium position. |
| Answer» SOLUTION :`(4PI)/(3)s` | |
| 3132. |
A thief intends to enter an apartment by climbing a ladder but foolishly places the upper end against a window. When he is 3.00 m up the ladder, the window is on the verge of shattering. His mass is 90.0 kg, the ladder's mass is 20.0 kg, the ladder's length is 5.00 m, and the foot of the ladder is 2.50 m from the base of the wall, on a non-slip ground surface. What are (a) the magnitude of the force on the glass from the ladder, (b) the magnitude of the force on the ladder from the ground, and (c) the angle between that ground force and the horizontal? |
| Answer» Solution :(a) 362 N (INWARD), (b) `1.14 XX 10^(3)N`, (c) `71.4^(0)` | |
| 3133. |
Draw phasor diagram for the ac circuit comprising of a pure inductor. |
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Answer» Solution :`epsilon = epsilon_(0) SIN omega t` `epsilon -L (d i)/(d t) = 0` `(d i)/(d t) = (epsilon)/(L) = (epsilon_(0))/(L) sin omega t` `underset(0)overset(i)int (di)/(d t) d t = (epsilon_(0))/(L) underset(0)overset(i)int sin omega tdt` `i= -(epsilon_(0))/(omega L) cos omega t` `rArr i = - (epsilon_(0))/(X_(L)) sin (omega t - pi//2), sin C^(e)(X_(L) = omega L)` `rArr i = i_(0) sin(omega t - (pi)/(2))` where , `i_(0) = (-epsilon_(0))/(X_(L))` 
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| 3134. |
A battery of internal resistance 4Omega is connected to the network of resistances as shown in the fig. In order that the maximum power can be delivered to the network, the value of R in Omega should be |
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Answer» `4//9` |
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| 3135. |
Charge q is uniformly directed over a thin half ring of radius R. The electric field at the centre of the ring is |
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Answer» `Q/(2 PI^2 epsilon_2 R^2)` |
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| 3136. |
A spaceship whose rest length is 280 m has a speed of 0.94c with respect to a certain reference frame. A micrometeorite, also with a speed of 0.94c in this frame, passes the sapceship on an antiparallel track. How long does it take this object to pass the ship as measured on the ship? |
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Answer» |
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| 3137. |
If m and M are the masses of two bodies that are tied at two ends of a meter scale that is balanced on a sharp edge of a heavy board wedge. If M=20g, its distance from centere = 30 cm and distance of mass m from centre is 25cm when metre scale is balanced, then m is |
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Answer» 23g |
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| 3138. |
The depletion layer in the p-n junction region is caused by |
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Answer» drift of holes |
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| 3139. |
When a carpet is beaten with a stick, dust comes outof it. Explain. |
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Answer» NEWTON's IST LAW of motion |
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| 3140. |
The temperature of a black body is 3000K. When the black body cools then at any time the wavelength of maximum energy density has changed by Delta lambda = 9 mu m . The temperature of black body at that time will be - |
| Answer» ANSWER :A | |
| 3141. |
In order to increase the sensitivity of a moving coil galvanometer, ______ should be decreased. |
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Answer» MAGNETIC field B `Phi/I=(NAB)/krArr(Phi/I)prop1/k` `rArr` By decreasing torsional constant of a spring, we can increase current sensitivity. |
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| 3142. |
The work done in rotating a dipole from the parallel to the perpendicular direction with the electric field is |
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Answer» 2PE |
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| 3143. |
कुल से निम्न स्तर का वर्गिको संवर्ग है : |
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Answer» वर्ग |
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| 3144. |
Co-efficient of absorption is 0.6 for a thermos body and amount of heat incident on itis 100J, then heat transmitted thought it is |
| Answer» Answer :D | |
| 3145. |
The electric field in a region is given by : E = (4axysqrt(z))hat(i)+(2ax^(2)y//sqrt(z))hat(j)+(ax^(2)y//sqrt(z))hat(k) , where a ispositive constant. The equation of an equipotential surface will be of the form |
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Answer» z = CONSTANT /`[x^(2)y^(2)]` |
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| 3146. |
Two satellites of same are orbiting round the earthh at heights of r_(1) and r_(2) from the centre of earth. Their potential energies are in the ratio of |
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Answer» `r_(2)//r_(1)` |
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| 3147. |
The correct curve between the stopping potential (V_(0)) and intensity of incident light (I) is |
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Answer»
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| 3148. |
A piece of wire carrying a current of 6A is bent in the form of a circular arc of radius 10cm and it subtends an angle of 120^@ at the centre. Find the magnetic field B due to this piece of wire at the centre. |
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Answer» `0.5 XX 10^(-5) T ` |
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| 3149. |
Explain the mutual induction between two long solenoids. Obtain an expression for the mutual inductance. |
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Answer» Solution :i. `S_(1)andS_(2)` are TWO LONG solenoids each length l. The solenoid `S_(2)` is wound closely over the solenoid `S_(1).` ii. `N_(1)andN_(2)` are the number of turns in the solenoids `S_(1)andS_(2)` respectively. Both the solenoids are considered to have the same area of cross section A as they are closely wound together.  iii. `I_(1)` is the current flowing through the solenoid `S_(1).` The magnetic field `B_(1)` PRODUCED at any point inside the solenoid `S_(1)` due to the current `I_(1)` is `B_(1)=mu_(0)(N_(l))/(l)I_(1)""......(1)` iv. The magnetic flux linked with each turn of `S_(2)` is EQUAL to `B_(1)A.` Total magnetic flux linked with solenoid `S_(2)` having `N_(2)` turns is `phi=B_(1)AN_(2)` Substituting for `B_(1)` from equation (1) `phi_(2)=(mu_(0)N_(1)N_(2)AI_(1))/(l)""....(2)` But, `phi_(2)=MI_(1)"".......(3)` where M is the coefficient of mutual induction between `S_(1)andS_(2).` From equtions (2) and (3), `ML_(1)=(mu_(0)N_(1)N_(2)AI_(1))/(l),M=(muN_(1)N_(2)A)/(l)` v. If the core is filled with a magnetice material of permeabilty `mu,` `M=(muN_(1)N_(2)A)/(l)` |
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| 3150. |
Current provided by a battery is maximum when |
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Answer» Internal RESISTANCE EQUAL to external resistance |
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