Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

The value of capacitive reactance is given by ……

Answer»

`X_C=-1/(OMEGAC)`
`X_C=1/(omegaC)`
`X_C=-j/(omegaC)`
`X_C=1/sqrt(omegaC)`

ANSWER :B
2.

An electric kettle has two coils when one of the coils is switched on, the kettle can boil water inside it in 12 minutes to boil. When the coil are connected in series

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ANSWER :36 MIN
3.

Trackof three chargedparticlesin a uniform electroastatic field give the sign of the three charges which particle has the highest charge to mass ratio

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SOLUTION :`5.7 xx10^(-3) N`
4.

Consider the situation of the previous problem. A charge of 2.0xx 10 ^(-4) C is moved from the point A to the point B. find the change in electrical potential energy U_B-U_A for the cases (a) , (b) and ( c ).

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SOLUTION :(a) the electric field is ALONG x DIRECTION Thus potential DIFFERENCE between `(0,0) and (4,2) is, `
` dV=-edx=-20xx4 `
` =-80volt `
` Potential energy (U_B-U_A) between the points=dVxxq `
` =-80xx-2.0xx10^-4 `
` =160xx10^-4=0.016J `
` (b) A=(4m,2m),B=(6m,5m) `
` dV=-exxdx=-20xx2=-40v `
` Potential energy (U_B-U_A) between the points=dVxxq `
` =-40xx(-2xx10^-4) `
` 80xx10^-4=0.008J `
` (c) A=(0,0) B=(6m,5m) `
` dV=-E.dx=-20xx6`
` -120V `
` potential energy (U_B-U_A) between the points A and B=dVxxq `
` =-120xx(-2xx10^-4) `
` 240xx10^-4=0.024J `
5.

What are mirage and looming?

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Solution :MIRAGE : Miragetakes PLACE in hot regions. Thelightform distanceobjects apperars to be REFLECTED from ground . Formirage to from refractive INDEX goes on increasingas we goup.
Looming : Loomingtakes placein coldregion. Thelight from distantobject APPEARS to be flying. For lomming toform refractive indexgoeson decreasing ,
6.

In a semiconductor holes are as mobile as electrons.

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Solution :FALSE - In a semiconductors HOLES are LESS mobile than ELECTRONS.
7.

Calculate the charge storing capacity of a pair of conductors if the potential differencebetween them is 12 V and when one trillion electrons are transferred from one conductor to the other.

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SOLUTION :`1.34xx10^(-8)F`
8.

The direction of magnetic induction at any point along the line making theta with a short dipole axis is :

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`PHI = tan^(-1) [1/4 tan THETA ]`
`phi = tan^(-1) [1/2 tan theta ]`
`phi = tan^(-1) [ 2 tan theta]`
`phi = tan^(-1) [ tan theta]`

Answer :B
9.

Ifmu_k is the coefficient of kinetic friction, mu_r the coefficient of rolling friction and mu_s the coefficient of static friction, then

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`mu_sgtmu_kgtmu_r`
`mu_sgtmu_kgtmu_r`
`mu_sgtmu_kgtmu_r`
`mu_sgtmu_rgtmu_k`

ANSWER :A
10.

When did Kailash Satyarthi win Nobel Prize :

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10TH OCTOBER 2014
20th October 2014
19th NOVEMBER 2014
None of the above

Answer :A
11.

(a) Magnetic field lines show the direction (at every point) along which a small magnetised needle aligns (at the point). Do the magnetic field lines also represent the lines of force on a moving charged particle at every point ?

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Solution :(a) Magnetic FIELD lines do not represent lines of force because magnetic force exerted on a charged particle is not acting along the field lines but it is ALWAYS DIRECTED perpendicular lines but it is always directed perpendicular by charged particle is different than the field LINE.
12.

1000 identical drops of mercury are charged to potential of 1V each. They join to form a single drop. The potential of this drop will be

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`0.01V`
`0.1V`
`10V`
`100V`

ANSWER :D
13.

A potentiometer wire has length 4 m and resistance 8Omega. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient 1 mV per cm on the wire is :

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`32 Omega`
`40 Omega`
`44 Omega`
`48 Omega`

Solution :`32 Omega`
Potential gradient, `sigma = (1mV)/(cm)`
= ` (10^(-3))/(10^(-2) m) =0.1 (V)/(m)`
Suppose, resistance R is CONNECTED in serieswith POTENTIOMETER.
`therefore ` Current in circuit I = `(V)/(8 + R) = (2)/(8 + R) `
Potential gradient , `sigma = I rho `
`0.1 = (2)/(8+ R) xx (8)/(4)`
` [ because gamma = ("resistance of wire")/("LENGTH of wire") ] `
8 + R = `(4)/(1)`
`therefore R = 40 - 8`
`therefore R= 32 Omega`
14.

Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism of given prism. Then the angle of prism is ...... (sin 48^@36. = 0.75)

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`41^@24.`
`60^@`
`80^@`
`82^@48.`

Solution :`eta=(SIN((A+delta_m)/(2)))/(sin""A/2)=(sinA)/(sin""A/2)[because delta_m=A]`
`1.5=(2sin""A/2os""A/2)/(sin""A/2)`
`0.75=cos""A/2`
`0.75=sin(90^@-A/2)`
`therefore 48^@36.=90^@-A/2`
`therefore A/2=90^@-48^@36.`
`therefore A/2=41^@24.`
`therefore A=82^@48.`
15.

Two linear paraller conductors caryyng currents in the same direction …………………each other but conductors carrying currents in mutually opposite directions …………………. .

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SOLUTION :ATTRACT, REPEL
16.

The statement that is not correct for the periodic classification of elements is :

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the properties of elements are the periodic functions of their atomic numbers.
non`-` metallic elements are lesser in number than metallic elements
the first ionisation energies of elements along a period do not vary in a REGULAR manner with increase in atomic number
for transition elements the `d-` SUBSHELLS are filled with electrons monotonically with increase in atomic number.

Solution :The `d-` SUB shells are not filled with electrons monotonically with increase in atomic number. There are some EXCEPTIONS like `Cr,Cu ` ETC.
17.

There are two groups of mutually synchronized clocks K and K^' moving relative to each other with a velocity v as shown in Figure. The moment when the clock A^' gets opposite the clock A is taken for the beginning of the time count. Draw the approximate position of hands of all the clocks at this moment "in terms of the K clocks", "in terms of the K^' clocks".

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Solution :By Lorentz transformation `t^'=(1)/(sqrt(1-v^2/c^2))(t-(vx)/(c^2))`
So at TIME `t=0`, `t^'=(vx)/(c^2)(1)/(sqrt(1-v^2//c^2))`
If `xgt0t^'lt0`, if `xlt0`, `t^'gt0` and we GET the diagram given below "in terms of the K-clock".

The situation in terms of the `K^'` clock is reversed.
18.

Give the ratio of velocities of infrared and ultaviolet rays in vacuum.

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SOLUTION :RATIO = 1
19.

Define self-inductance and give its unit. Write down the expression for the self-inductance of a long solenoid of length l having N-turns.

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Solution :For definition of self-inductance, see Point Numbers 16 and 17 under the heading "Chapter At a Glance". SI unit of self-inductance is HENRY (H).
The self-inductance of an air core long SOLENOID of length I, area of cross-section A and having N-turns in all its GIVEN by `L = (mu_(0) N^(2)A)/l`
If a magnetic material having relative permeability `mu_(r)` is inserted in the core then self-inductance will be
`L = (mu_(0)mu_(r) N^(2)A)/l.`
20.

One plate of a capacitor is connected to a spring as shown. Area of both the plate is A and separation is d when capacitor is uncharged. When capacitor is charged, in steady state separation between the plates is 0.8 d. The force constant of the spring is

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`(epsi_0 AE^(3))/( 2d^3)`
`(65)/(32)(epsi_0AE^(2))/( d^3)`
`(125)/( 16)( epsi_0 AE^2)/(d^3)`
`(125)/(32)(epsi_0AE^(2))/( d^3)`

ANSWER :D
21.

An object is placed on the surface of smooth inclined plane of inclination . It takes time to reach the bottom. If the same object is allowed to slide down the rough inclined plane of the same inclination, the time to reach the bottom is increased n times, where n gt 1. The coefficient of friction for the plane is :

Answer»

`mu=tantheta[1-(1)/(n^(2))]`
`mucostheta[1-(1)/(n^(2))]`
`mu=tantheta[1-(1)/(n^(2))]^((1)/(2))`
`mu=costheta[1-(1)/(n^(2))]^((1)/(2))`

Solution :Here acceleration down the smooth plane is `g sin theta` and the rough inclined plane is `(g sin - mug COS theta)` If "l" is the length of plane, then
`l=(1)/(2)gsin theta xxt^(2)`
and alos `l=(1)/(2)(sin theta-mu cos theta)gn^(2)t^(2)`
or`SINTHETA=(sin theta-mu cos theta)n^(2)`
or `sin theta =sin thetan^(2)-mun^(2)-mun^(2)cos theta`
or `mun^(2) cos theta= sin theta(n^(2)-1)` or` mu=tan theta[1-(1)/(n^(2))]`
`(a) is the CHOICE
22.

A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet. (i) In which direction will it move? (ii) What will be the direction of its magnetic moment?

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Solution :(i) The BALL of superconducting MATERIAL behaves as a perfect diamagnet and therefore MOVES away from the bar magnet.
(ii) The direction of INDUCED magnetic moment DEVELOPED in ball is opposite to the direction of magnetic moment of bar magnet.
23.

Show the variation of the emf generated versus time as the armature is rotated with respect to the direction of the magnetic field.

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SOLUTION :
24.

The value of ground state energy of hydrogen atom is - 13.6 eV. (a) Find the energy required to move an electron from the ground state to the first excited state of the atom. (b) Determine (i) the kinetic energy, and (ii) the orbital radius in the first excited state of the atom. (Giventhe valueof Bohr radius= 0.53Å)

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Solution :(a) `because` Ground state energy of electron (N = 1) is E = - 13.6 eV
`therefore ` Energy for firstexcited state `(n =2) E_(2) =(-13.6)/(n^(2)) = (-13.6)/((2)^(2)) = - 3.4 eV`
`therefore ` Energyrequired to be given to electron `Delta E = E_(2) - E_(1) = - 3.4 - (-13.6) = + 10.2 eV`
(b) (i) Kineticenergyof electron in 1st excited state `K_(2) = -E_(2) = + 3.4 eV`
(ii) Orbital RADIUS in first excited state `r_(2) = n^(2) a_(0) = (2)^(@) xx 0.53 Å = 2.12 Å`
25.

The vertical component of earth's magnetic field is zero at a place where angle of dip is :

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`0^(@)`
`45^(@)`
`60^(@)`
`90^(@)`

ANSWER :A
26.

Two resistive solenoids are made of same geometry and same number of turns per unit length. Copper wire used to make the solenoids are having different radius of cross section. Which of the following quantities are different for solenoids?

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Time constant if both the SOLENOIDS are connected to identical batteries
Rate of joules heating when same current is passed through the solenoids
Magnetic field energy stored in solenoids when same current is passed through the solenoids
Self-inductance

Solution :Self-inductance is independent of radius of wire used to make it. Hence self-inductance for both is the same, hence option (d) is wrong. Energy stored in the 1 solenoid is given by `U = (1)/(2)LI^(2)` hence energy stored in 2 both the solenoids is going to be same when same current is passed through both the solenoids. Hence, option (c) is wrong. Radius of cross section of wire affects the resistance of COIL and hence it affects time constant of the circuit and rate of heat dissipation when same current is passed through them. Hence, options (a) and (b) are CORRECT.
27.

A parallel plate capacitor of capacitance 40 muF is charged to a potential difference of 250 V and then the battery is disconnected. Separationbetween the paltes of the capacitoris 5 mm. A dielectric slab of dielectric constant 5 and thickness 3 mm is introducedin -between the plates of capacitor. Calcualte final charge and potential difference between the plates. Also calculate energy loss due to the introduction of dielectric slab.

Answer»

Solution :Given, initial capacitance of the parallel plate CAPACITOR, `C_(0)=(epsilon_(0)A)/(d)=40muF`
Initial potential difference between the plates of capacitor, `V_(0)=250V`
Separation between the plates of capacitor,
`d=5 mm = 5xx10^(-3)m`
Initial charge stored in capacitor,
`Q_(0)=C_(0)V_(0)=40xx10^(-6)xx250`
`Q_(0)=10^(4)muC`
On inserting the DIELECTRIC of thickness t and dielectric constant K,
Now the new capacitance will be:
`C.=(epsilon_(0)A)/((d-t+(t)/(K)))`
`C.=(epsilon_(0)A)/((d-3+(3)/(5))xx10^(-3))`
`=(epsilon_(0)A)/((5-3+(3)/(5))xx10^(-3))`
`=(epsilon_(0)A)/((13)/(5)xx10^(-3))`
We know that, `(epsilon_(0)A)/(5xx10^(-3)) =40muF`
`therefore C.=(((epsilon_(0)A)/(5xx10^(-3)))5xx10^(-3))/((13)/(5)xx10^(-3))`
`rArr C.=((epsilon_(0)A)/(5xx10^(-3)))xx(5xx10^(-3)xx5)/(13xx10^(-3))`
`=(25)/(13)xx40 muF`
`C.=76.92muF`
Let the new potential difference be V. across plates.
New charge on capacitor will be same as before.
`therefore V.C.=Q_(0)`
`V.=(C_(0)V_(0))/(C.)`
`V.=(40xx10^(-6))/(76.92xx10^(-6))xx250`
`V.=130V`
Initially, energy stored in the capacitor is:
`U=(1)/(2)CV^(2)`
`=(1)/(2)xx40xx10^(-6)xx(250)^(2)=1.25J`
Finally, energy stored in the capacitor is : `U.=(1)/(2)C.(V.)^(2)`
`=(1)/(2)xx76.92xx(130)^(2)=0.65J`
Energy loss due to introduction of dielectric slab
`=U-U.=1.25-0.65=0.60J`
28.

A circular racetrack of radius 100 m is banked at an angle of 45^(@). Whatis the (i) Optimum speed of race car to avoid wear and tear of its tyres? (ii) Maximum permissible speed to avoid slipping if the coefficient of friction is 0.2?

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Solution :At the optimum speed, optimum speed `v_(0)` is given by EQUATION
`v_(0)=(Rg tan THETA)^(1//2)`
`=(100xx9.8xx tan45)^(1//2)`
`=31.3 m//s`
The MAXIMUM PERMISSIBLE speed `v_("MAX")` is given by
`v_("max")=("Rg"(mu_(s)+tan theta)/(1-mu_(s)tan theta))^(1//2)`
`=(100xx9.8xx (0.2+tan45)/(1-0.2 tan45))^(1//2)`
`=38.34 m//s`
29.

The carbon resistor has three orange bands. The maximum value of resistance offered by the resistor will be ..... .

Answer»

49.6 k`Omega`
`39.6 k Omega`
`33 k Omega`
`26.4 k Omega`

Solution :`39.6 k Omega`
Maximum resistance = 20%of R+ R
= ` 33 XX 10^(3) + (20 xx 33 xx 10^(3))/(100) `
= 33000 + 6600
= 39600 `Omega`
` = 39.6 xx 10^(3) Omega = 39.6 k Omega`
30.

A boy of mass 60 kg standing on a plat-form of mass 40 kg placed over a smooth horizontal surface. He throw, a stone of mass 1 kg with velocity u - 10 m/s at on angle of 45° w.r.t the ground. The displacement of platform (with boy) on the horizontal surfaces when the stone lands on the ground is (g = 10 m//s^(2))

Answer»

Solution :`1 xx u COS 45^(@) =(60 + 40) v, v=1/(10sqrt(2))`
TIME of flight, `T_(f) =(2u SIN theta)/G = sqrt(2)` sec.
`S = v xx T_(f) =0.1` m
31.

At what angle with the horizontal a player should throw a ball so that it may go to a maximum distance ? For maximum height ?

Answer»

SOLUTION :For MAXIMUM DISTANCE, the ANGLE of projection should be `45^@` while for maximum HEIGHT, the angle of projection should be `90^@`
32.

The velocity of a particle is v=6hat(i)+2hat(j)-2hat(k).The component of the velocity parallel to vectora=hat(i)+hat(j)+hat(k)in vector form is

Answer»

`6HAT(i)+2hat(J)+2hat(k)`
`2hat(i)+2hat(j)+2hat(k)`
`hat(i)+hat(j)+hat(k)`
`6hat(i)+2hat(j)-2hat(k)`

Solution :(B) Magnitude of component of v along a
`=(v.a)/(a)=(6+2-2)/(sqrt(3))=2sqrt(3)`
`"Now"hat(a)=(hat(i)+hat(j)+hat(k))/(sqrt(3))`
:.Component in VECTOR form
`=2sqrt(3)hat(a)`
`=(2hat(i)+2hat(j)+2hat(k))`
33.

Feeling of weightlessness in a satellite or spaceship is due to

Answer»

ABSENCE of inertia
absence of accelerating force
acceleration of SATELLITE
FREE fall of spaceship.

Solution :Weightlessness in a satellite or spaceship is due acceleration of satellite as whole of GRAVITYIS used in providing this acceleration.
Thus correct choice is (c ).
34.

Calculate the emitter current in a transistor for which beta = 50 , base current I_(B) = 25 mu A .

Answer»

SOLUTION :Data SUPPLIED ,
`beta = 50 , I_(B) = 25 MU A = 0.025 mA , beta = (I_(C))/(I_(B))`
`I_(C) = beta I_(B) = 50 xx 0.025 = 1.25` mA
`I_(E) = I_(B) + I_(C) = 0.025 + 1.25 = 1.275` mA
35.

A hollowing insulated conducting sphere is given a positive charge of 10mu what will be the electric field at the center of the sphere if its radius is 2 meters?

Answer»

`20 MU Cm^-2`
5 Cm
zero
8 Cm

Answer :C
36.

A vector barQ which has a magnitude of 8 is added to the vector barP which lies along the X-axis. The resultant of these two vectorsis a third vector barRwhich lies along the Y-axis and has a magnitude twice that of barP. The magnitude of barP is

Answer»

SOLUTION :`(8)/(SQRT(5))`
37.

Two long thin wires ABC and DEF are arranged as shown. They carry currents I as shown . The magnitude of the magnetiofield at O is :

Answer»

zero
`(mu_(0)I)/( 4 pi a)`
`(mu_(0) I)/(4 pi a)`
`(mu_(0) I)/( 2 sqrt(2 pi a))`

ANSWER :C
38.

Consider a hypothetical case of a radinuclide A which produces a daughter B which is turn decays to a stable product C. the half life of B is 5 times that of A. take initial population of both B and C to be zero. (a) Using qualitative arguments only, plot the variation of population of A,B and C with time. (b) Find the ratio of population of B to that of A at the instant the population of B becomes maximum. (c) Describe qualitatively how the slope of the graph of population oc C changes with time.

Answer»


ANSWER :`(##IJA_PHY_V02_C16_E01_053_A01##)`
(b) `(N_(B))/(N_(A))=5`
(C) The slope FIRST INCREASES. It becomes maximum when `N_(B)` is maximum. Thereafter the slope decreases.
39.

Match the statements in Column I with the statements in Column II . One or more than one choice from Column II can match with a statement from Column I.

Answer»

<P>

SOLUTION :`(a) to (r) , (b) to (Q), (c) to (p ) , (d) to (s) `
40.

One end of a 60m long rope of mass 1.8 kg is tied to a rigid horizontal support held high above the ground. The rope hangs vertically and kept taut by a weight eW suspended at its lower end. A person jerks the lower end of the rope sideways in a sinusoidal manner and a transverse wave of frequency 2 Hz and amplitude 10cm passes along the rope such that there are 2 cycles of the wave in the total length of the rope. Neglecting the weight of the rope as compared to the suspended weight W and with g = 10m//s^2, answer the following questions. (S = 75 xx l0^(-4) m^(2)), In the questions above, weight of the rope has been neglected as compared to the suspended weight. However, if we also account for the weight of the rope, the speed of the wave at the top of the rope will be

Answer»

60 m/s
64.8 m/s
62.6 m/s
68.4 m/s

ANSWER :B
41.

When we look at the sky in a clear night, we find the sky full of stars, filling every. portion of the sky almost uniformly. But on a very careful observation through highly sophisticated instruments, one would observe dark areas in portions of the observable universe devoid of stars. Such a region is known as a dark nebula. A dark nebula is basically a cold gas cloud in interstellar space that contains enough material to block out light from stars behind it. A typical dark nebuta is about 20 light years in diameter and contains about 50 hydrogen atoms per cubic centimeter (monatomic hydrogen, not H_2) at a temperature of about 20 K. (A light year is the distance light travels in vacuum in one year and is equal to 9.46 xx 10^(15)m). What would be the mean free time (average time between collisions)

Answer»

`6.4 xx 10^(8)s`
`3.2 xx 10^(8)s`
`6.4 xx 10^(9)s`
`3.2 xx 10^(7)s`

ANSWER :A
42.

When we look at the sky in a clear night, we find the sky full of stars, filling every. portion of the sky almost uniformly. But on a very careful observation through highly sophisticated instruments, one would observe dark areas in portions of the observable universe devoid of stars. Such a region is known as a dark nebula. A dark nebula is basically a cold gas cloud in interstellar space that contains enough material to block out light from stars behind it. A typical dark nebuta is about 20 light years in diameter and contains about 50 hydrogen atoms per cubic centimeter (monatomic hydrogen, not H_2) at a temperature of about 20 K. (A light year is the distance light travels in vacuum in one year and is equal to 9.46 xx 10^(15)m). Whal would be the pressure inside a dark nebula?

Answer»

`1.4 XX 10^(-12)Pa`
`2.8 xx 10^(-14)Pa`
`2.8 xx 10^(-12)Pa`
`1.4 xx 10^(-14)Pa`

ANSWER :D
43.

One end of a 60m long rope of mass 1.8 kg is tied to a rigid horizontal support held high above the ground. The rope hangs vertically and kept taut by a weight eW suspended at its lower end. A person jerks the lower end of the rope sideways in a sinusoidal manner and a transverse wave of frequency 2 Hz and amplitude 10cm passes along the rope such that there are 2 cycles of the wave in the total length of the rope. Neglecting the weight of the rope as compared to the suspended weight W and with g = 10m//s^2, answer the following questions. (S = 75 xx l0^(-4) m^(2))), Average power will be nearly

Answer»

3.2 w
2.25 w
1.4 w
0.75 w

Answer :A
44.

When we look at the sky in a clear night, we find the sky full of stars, filling every. portion of the sky almost uniformly. But on a very careful observation through highly sophisticated instruments, one would observe dark areas in portions of the observable universe devoid of stars. Such a region is known as a dark nebula. A dark nebula is basically a cold gas cloud in interstellar space that contains enough material to block out light from stars behind it. A typical dark nebuta is about 20 light years in diameter and contains about 50 hydrogen atoms per cubic centimeter (monatomic hydrogen, not H_2) at a temperature of about 20 K. (A light year is the distance light travels in vacuum in one year and is equal to 9.46 xx 10^(15)m). If the space around the nebula were a vacuum then

Answer»

it WOULD REMAIN stable
It would evaporate
It would COLLAPSE into itself
it would MAKE a hollo

Answer :B
45.

a. Give two radioacitve elements which are not formed in observable quantities in nature. b. Give reason for this.

Answer»

Solution :a. PLUTONIUM, TRITIUM
b. They have very SMALL half-life period. Hence the REASON.
46.

In the above question, if the velocity of the object on reaching the bottom is upsilon for the smooth plane and(upsilon)/(n) for the rough plane, then the coefficient of friction is given by :

Answer»

`mu=tantheta[1-(1)/(n^(2))]`
`mu=costheta[1-(1)/(n^(2))]`
`mu=tantheta[1-(1)/(n^(2))]^((1)/(2))`
`mu=costheta[1-(1)/(n^(2))]^((1)/(2))`

Solution :Applying `upsilon^(2) - u^(2)= 2as` we have `upsilon^(2) = 2G sin thetaxxl`or `upsilon^(2) = 2g sin thetaxx L` and`(upsilon^(2))/(n^(2))=2g(sintheta-mucostheta)l` on
solving we GET the same value of .u. as in the above QUESTION (a) is the choice
47.

One end of a 60m long rope of mass 1.8 kg is tied to a rigid horizontal support held high above the ground. The rope hangs vertically and kept taut by a weight eW suspended at its lower end. A person jerks the lower end of the rope sideways in a sinusoidal manner and a transverse wave of frequency 2 Hz and amplitude 10cm passes along the rope such that there are 2 cycles of the wave in the total length of the rope. Neglecting the weight of the rope as compared to the suspended weight W and with g = 10m//s^2, answer the following questions. (S = 75 xx l0^(-4) m^(2))), Mass corresponding to the suspended weight W is

Answer»

14.2kg
10.8kg
6.6kg
8.4k

Answer :B
48.

A small ball of mass 1 kg and charge (2)/(3)mu C is placed at the centre of a uniformly charged sphere of radius 1 m and charge (1)/(3)mC. A narrow smooth horizontal groove is made in the sphere from centre to suface as shown in figure. The sphere is made to rotate about its vertical diameter at a constant rate of (1)/(2pi) revolutions per second. Find the speed w.r.z ground (inm/s) with which the ball slides out from the groove. Neglect any magnetic force acting on ball.

Answer»


ANSWER :B
49.

In case of a semi conductor, which one of the following statements is wrong?

Answer»

Resistivity is in between that of a CONDUCTOR and insulator
Temperature coefficient of RESISTANCE is negative
Doping increases conductivity
At absolute zero it BEHAVES as a conductor

Answer :D
50.

(A):The body falls freely,when acceleration of body is equal to acceleration due to gravity. (R ):A body falling freely,when acceleration of body is equal to acceleration due to gravity. (R ):A body falling freely will have constant velocity.'

Answer»


ANSWER :C