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2101.

A ball is projected from the bottom of an inclined plane of inclination 30^@, with a velocity of 30 ms^(-1), at an angle of 30^@ with the inclined plane. If g = 10 ms^(-2) , then the range of the ball on given inclined plane is

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12 m
60 m
120 m
600 m

Answer :B
Discussion
2102.

A 0.50 kg body oscillates in SHM on a spring that, when extended 2.0 mm from its equilibrium position, has an 8.0 N restoring force. What are (a) the angular frequency of oscillation, (b) the period of oscillation, and (c) the capacitance of an LC circuit with the same period if L is 5.0 H?

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ANSWER :(a)89 rad/s ; (b) `7.0xx10^(-2)` s ; (C) `2.5xx10^(-5)` F
Discussion
2103.

A cell placed in solution swells up. The solution is

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hypertonic
isotonic
hypotonic
both a and b

Answer :C
Discussion
2104.

The work done in increasing the size of soap film from 10cm xx 6cm to 10cm xx 11cm is 3 xx 10^-4 joule. The surface tension of the film is,

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`1.5 XX 10^-2 N/m`
`3.0 xx 10^-2 N/m`
`6.0 xx 10^-2 N/m`
`11.0 xx 10^-2 N/m`

ANSWER :B
Discussion
2105.

When the load resistance across a solar cell is zero, the current in the external circuit passed by the solar cell is called

Answer»

the open-circuit CURRENT
the REVERSE SATURATION current
the short-circuit current
the photocurrent.

Answer :C
Discussion
2106.

A star can be considered as a shpherical ball of hot gas of radius R.Inside the star the density of the gas is rho_r at a radius r and mass of the gas within this region is M_r.The correct differential equation for variation of mass with respect to radius ((dM_r)/(dt)) is (refer to the adjacent figure) A star in its prime age is said to be under equilibrium due to gravitational pull and outward radiation pressure (rho).Consider the shell of thickness dr .If the pressure on this shell is dp then the correct equation is (G is universal gravitational constant)

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ANSWER :`(DP)/(DR)=-(GM_r)/r^2`
Discussion
2107.

A star can be considered as a shpherical ball of hot gas of radius R.Inside the star the density of the gas is rho_r at a radius r and mass of the gas within this region is M_r.The correct differential equation for variation of mass with respect to radius ((dM_r)/(dt)) is (refer to the adjacent figure) In astronomy order of magnitude estimation plays an important role. The derivative (dp)/(dr) can be taken difference ratio (DeltaP)/(Deltar) . Consider the star has a radius R, pressure at its centre is P_C and pressure at outer layer is zero if the average mass is M_0/2 and average radius R_0/2 then the expression for P_C is

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ANSWER :`P_C=(GM_0^2)/(R_0^4)XX(3/(2PI))`
Discussion
2108.

A star can be considered as a shpherical ball of hot gas of radius R.Inside the star the density of the gas is rho_r at a radius r and mass of the gas within this region is M_r.The correct differential equation for variation of mass with respect to radius ((dM_r)/(dt)) is (refer to the adjacent figure) The value of mass and radius of sun are given by M_0=2xx10^30 kg and R_0=7xx10^5 km respectively.The pressure at the centre is about (G=6.67xx10^(-11) m^3 kg^(-1) s^(-2)).

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ANSWER :`5xx10^14 N//m^2`
Discussion
2109.

A star can be considered as a shpherical ball of hot gas of radius R.Inside the star the density of the gas is rho_r at a radius r and mass of the gas within this region is M_r.The correct differential equation for variation of mass with respect to radius ((dM_r)/(dt)) is (refer to the adjacent figure) Assuming that the gas inside the sun behaves very much like the perfect gas, the temperature at the centre of the sun is nearly (the number density of gas particles =(2rho)/(M_H) , Boltzmann constant K_B=1.4xx10^(-23) J.K^(-1) and mass of proton M_H=1.67xx10^(-27) kg)

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ANSWER :`2.10xx10^7`
Discussion
2110.

An inductor 200 mH, a capacitor C and a resistor 10 ohm are connected in series with a 100 V, 50 s^(-1) a.c. source. If the current and voltage are in phase with each other, calculate the capacitance of the capacitor.

Answer»

Solution :Here, L = 200 mH= 0.2 H, `V_(rms) = 100 V, v= 50 s^(-1)` or `omega = 2PI v = 2 XX 3.14 xx 50 = 314 rad s^(-1)`
As V and I are in phase, hence `X_(L) = X_(C)`
`rArr C = 1/(L omega^(2)) = 1/(2 xx (314)^(2)) = 5.07 xx 10^(-5) F = 50.7 muF`.
Discussion
2111.

The angle of contact for liquid which completely wets the solid surface is

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`0^@`
`90^@`
More than `90^@`
LESS than `90^@`

ANSWER :A
Discussion
2112.

Figure shows a metre bridge in which a known resistance R and an unknown resistance X connected to it. The length of the total wire AB is 100cm. Then the balancing length for which the percentage error in measuring the value of x is minimum (neglect end corrections)

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0 cm
33.3 cm
50 cm
100 cm

Answer :C
Discussion
2113.

Metals are good conductors of heat as compared to the insulators because

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they CONTAIN free electrons
their ATOMS are RELATIVELY far apart.
their atoms collide frequently.
they have reflecting surfaces.

Answer :A
Discussion
2114.

Assuming the expression for the path difference between two interfering light waves for bright and dark fringes, obtain an expression for the fringe width. The central fringe of a double-slit interference pattern produced by light of wavelength 6000 Å shifts to the position of the 5th bright fringe on introducing a thin glass plate of refractive index 1.5 in front of one of the slits. Calculate the thickness of the plate.

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ANSWER :`6 MU m`
Discussion
2115.

A concave mirror of focal length f produces an image n times the size of the object .If the image is real, what is the distance of the object from the mirror ?

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`[(N+1)/n]F `
`[n/n+1]f`
`n^2f`
`n^2/f`

ANSWER :A
Discussion
2116.

Why are hole carriers present in n-type semiconductor ?

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SOLUTION :Due to the addition of pentavalent IMPURITY, n- type MATERIAL has a large number of free electrons. However, even at ROOM TEMPERATURE, some of the .of free .electrons and holes. Therefore, an n-type material has a large number of free electrons. and a small number of holes.
Discussion
2117.

A circular coil of radius 10 cm and 100 turns carries a current 1 A. What is the magnetic moment of the coil ?

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`3.142 xx 10^(4) Am^(2)`
`10^(4) A m^(2)`
`3.142 A m^(2)`
`3 A m^(2)`

SOLUTION :`M = NIA = NI pi r^(2) = 10^(2) xx 1 xx 3.142 xx 10^(-2) = 3.142 Am^(2)`
Discussion
2118.

Two cells of emf 1.5 V and 2 V and internal resistance 1Omega and 2Omegarespectively are connected inparallel to pass a current in the same direction through an external resistance of 5Omega.(i) Draw the circuit diagram.(ii) Using Kirchhoff's laws, calculate the current through each branch of the circuit and potential difference across the 5Omegaresistor.

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Solution :(i) The circuit diagram has been SHOWN in Fig.

(ii) In mesh ACRDBA, applying Kirchhoff.s second LAW, we have
` - (I_1 + I_2) .5- I_1 .1 + 1.5 = 0`
`rArr6I_1 + 5I_2 = 1.5 `....(i)
and for mesh CRDC, we have
` - (I_1 + I_2) .5 - I_2 .2 + 2= 0`
` RARR 5I_1 + 7I_2 = 2`....(ii)
On solving (i) and (ii), we get `I_1 = 1/34 A` and `I_2 = 9/34 A`
Current through external `(5OMEGA)` resistor
`= I_1 + I_2 = 1/34 + 9/34= 10/34 A = 5/17 A`
` therefore `Potential difference across the 5`Omega`resistor
`V = (I_1 + I_2) R = 5/17 xx 5 = 25/17 V`
Discussion
2119.

A window whose area is 2m^2 opens on a street where the street noise result in .an intensity level at the window of 60 dB. If an acoustic absorber is fitted at the window, how much energy from street will it collect in five hours

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`26 XX 10^(-13) J`
`26 xx 10^(-10)` J
`36 xx 10^(-3) J`
`46 xx 10^(-10) J`

Answer :C
Discussion
2120.

A nichrome heating element connected to a 200 V supply draws an intial current of 2.2 A which settles down after a few seconds to a steady value of 2.0 A. Find the steady temperature of the heating element. The room temperature is 30^(@)C and the average temperature coefficient of resistance of nichrome is 1.7xx10^(-4) "per"^(@)C.

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ANSWER :`621^(@)C`
Discussion
2121.

Red light, howsoever bright it is, cannto produce the emission of electrons from a clean zinc surface, but even weak ultraviolet radiation can do so. Why? X-rays of wavelength lamda fall on photosensitive surface, emitting electrons. Assuming that the work function of the surface can be neglected, prove that the de-Broglie wavelength of electrons emitted will be sqrt((hlamda)/(2mc)).

Answer»

Solution :For photoelectric emission, the energy of incident photon MUST be greater than the work function of the METAL. For zinc work function is about 3.4 eV. As energy of red light photon is less hence photo-emission will not TAKE PLACE. However, having much smaller wavelength the energy of ultraviolet light photon is much higher and, consequently, it can cause emission of electrons, even if its intensity is weak.
If we neglect work function of the photosensitive surface, then kinetic energy of photo-electrons will be
`K=hv=(hc)/(LAMDA)`
`therefore`de-Broglie wavelength of these photoelectrons will be
`lamda_("electron")=(h)/(sqrt(2mK))=(h)/(sqrt(2m*(hc)/(lamda)))=sqrt((hlamda)/(2mc))`.
Discussion
2122.

How will the conductor, carrying current I_(0) rotate immediately after it is released in following three cases [consider magnetic force only] (a)Conductor carrying current I_(0) is placed symmetrically above poles of a fixed U shaped magnet (figure a). (b)Conductor carrying current I0 is placed symmetrically at a distance from a fixed current (I_(1)) carrying wire (fig. (b)) (c) An insulated circular current carrying wire is held fixed in vertical plane. Conductor carrying current I_(0) is in the shape of a circle of diameter nearly equal to that of the fixed insulated circle. The planes of the two circles are perpendicular to each other (fig. (c)) with xy as common diameter

Answer»


Answer :(a) Conductor will ROTATE in HORIZONTAL plane (b) Conductor will rotate so as to get parallel to FIXED wire (c) CIRCULAR conductor will rotate so as to set the current `(I_(0))` parallel to `(I_(1))`
Discussion
2123.

Electric field inside conductor is always

Answer»

Zero
Constant
PROPORTIONAL to r
Inversely proportional to r

Answer :A
Discussion
2124.

In AM wave , total power of side hands is given by

Answer»

`P_(s) = (E_(c)^(2))/(4R)`
`P_(s) = (4 E_(c)^(2))/(R)`
`P_(s) = (m^(2) E_(s)^(2))/(4R)`
`P_(s) = (E_(c)^(2))/(4M^(2) R)`

ANSWER :C
Discussion
2125.

A red hot steel needle is suspended along the North South direction and cooled. What happens

Answer»

Solution :A weak magnetic moment is induced PARALLEL to the EXTERNAL FIELD and the needle BECOMES a weak magnet.
Discussion
2126.

NOT QUESTION

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the surface MUST be frictionless
the FORCE of friction between the bodies is zero.
the magnitude of normal force equal that of friction
It is possible that the bodies are rough and they do not slip on each other.

ANSWER :B::D
Discussion
2127.

Yellow light is incident on a smooth surface of a block of dense flint glass for which the refractive index is 1.6640. Find the polarising angle and the angle of refraction.

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ANSWER :`59^(@),31^(@)`
Discussion
2128.

How does current in a pure capacitor vary in terms of phase angle with the voltage across it?

Answer»

Solution : If`V = V_(0) SIN omega t`
`q = CV = CV_(0) sin omega t`
`I = (DQ)/(d t) = omega CV_(0) cos omega t`
or `I = omega CV_(0) sin (omega t + (pi)/(2))`
So, the current leads the APPLIED VOLTAGE in phase by `pi//2`.
Discussion
2129.

Two plane mirrors are placed parallel to each other at a distance L apart. A point object ( is placed between them, at a distance L/3 from one mirror. Both mirrors form multiple images. The distance between any two images cannot be

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3L/2
2L/3
2L
none

ANSWER :A
Discussion
2130.

Find the total number of fringes formed on screen in the L loyd's mirror arrangement shown.

Answer»


Solution :
`(0.1)/h_(1)=50/50Rightarrowh_(1)=1mm`
`(L+h_(1))/(0.1)=80/20 RightarrowL+h_(1)=4mm RightarrowL=3mm`
`B=(lambdaD)/d=(5xx10^(-7)XX1)/(2xx10^(-3))=2.5xx10^(-4)` m
`N=L/B = (3xx10^(-3))/(2.5xx10^(-4))=300/25=12`
Discussion
2131.

A particle is moving in a circle in front of a plane mirror in the situation as shown in figure. The plane of motion of the particle is perpendicular to the plane of mirror. Then the motion of image of particle with respected to the particle is

Answer»

along a PARABOLA
oscillating normal to the mirror
oscillating parallel to the mirror
along a circle

ANSWER :B
Discussion
2132.

Two capacitors C_(1) = 4 muFand C_(2) = 2muFare charged to same potential V = 500 Volt, but with opposite polarity as shown in the figure. The switches S_(1) and S_(2)are closed.

Answer»

The potential difference across the two CAPACITORS are same and is GIVEN by `(500)/3V`
The potential difference across the two capacitors are same and is given by `(1000)/3V`
The ratio of final ENERGY to initial energy of the system is 1/9.
The ratio of final energy to initial energy of the system is 4/9.

Answer :A::C::D
Discussion
2133.

Value of dielectric strength for air is ...... Vm^(-1)

Answer»

`3XX10^(4) `
`3 xx10^(6)`
`6xx10^(3)`
`4 xx10^(3)`

Solution :KNOWLEDGE BASED question
Discussion
2134.

In an L-C oscillator circuit having a completely charged capacitor, with the passage of time ........

Answer»

the electric current increases gradually
the ENERGY of the circuit CONTINUOUSLY increases.
the energy of the circuit continuously decreases.
there is a CONTINUOUS absorption of the electromagnetic wave.

Solution :According to the formula `Q = Q_0sinomega_0t`, as the charge on the CAPACITOR decreases, the energy ASSOCIATED with it is also decreases.
Discussion
2135.

The dimensions of the quantity namely (mu_(0)ce^(2))/(2h) is : where mu_(0)- permebility of free space, c-velocity of light, e=electron charge and h being Plank's constant :

Answer»

`[M^(0)LT]`
`[M^(0)L^(0)T]`
`[M^(0)L^(0)T^(0)]`
`[M^(-2)L^(-1)T^(3)]`

Solution :Here LET `X=(mu_(0)CE^(2))/(2h)`
Putting the dimensions of various QUANTITIES we get
`X=[MLT^(-2)A^(-2)]xx[M^(0)L^(1)T^(-1)]xx[A^(2)T^(2)]`
`=[M^(0)L^(0)T^(0)]xx[M^(-1)L^(-2)T]`
Thus `(c )` is the correct CHOICE .
`:.` Correct choice is `(c )`
Discussion
2136.

When current changes from 3A to 2A in one millisecond a coil produces 5V. Its self inductance is

Answer»

5100H
5mH
50H
5H

Answer :B
Discussion
2137.

According to Max Planck's energy of photon is:

Answer»

E=hv
E=h
e=v
`h/delta`

ANSWER :A
Discussion
2138.

A potential difference of 6V is applied to a coil of inductance 0.5H and a resistance of 4 ohm connected in series. The time taken for the current to reach half the maximum value is (in seconds)

Answer»

`(log_e 2)/(8)`
`(log_e 2)/(4)`
`(log_e 4)/(8)`
`(log_e 4)/(4)`

ANSWER :A
Discussion
2139.

Explain the ray diagram for image obtained for concave mirror by considering three rays.

Answer»

Solution :
Figure shows the ray diagram considering three rays.
It shows the image A.B. of an object AB FORMED by a CONCAVE mirror.
Number of rays emanate from any SOURCE in all directions. Thus, point A. is image point of A if every ray ORIGINATING at point A and falling on the concave mirror after REFLECTION passes through the point A..
Discussion
2140.

Two wires of same metal have same length, but their cross sections are in the ratio 3 : 1. They are joined in series. The resistance of the thicker wire is 10 Omega The total resistance of the combination is

Answer»

`5/2 OMEGA`
`40/3 Omega`
`40 Omega`
`100 Omega`

ANSWER :A::D
Discussion
2141.

In Davisson-Germer experiment, an electron beam of energy 60 eV is incident normally on a crystal surface. The maximum intensity is obtained at an angle of 60^(@) from the direction of incident beam, then the inter atomic distance in the crystal lattice will be :

Answer»

`1.83Å`
`3.6 Å`
`18Å`
`0.18Å`

SOLUTION :`LAMBDA= d sin phi=(12*27)/(sqrt(V))`
`d=(12*27xx2)/(sqrt(3xx60))=1*83Å`
Discussion
2142.

A square wire of side 6.0 cm is placed 25.0 cm away from a concave mirror of focal length 10.0 cm. The area enclosed by the image of the wire, whose centre is on the principal axis, is :

Answer»

4 `cm^2`
8 `cm^2`
12 `cm^2`
16 `cm^2`

ANSWER :D
Discussion
2143.

A,B and C are three large, parallel conducting plates, placed horizontally. A and C are rigidly fixed and earthed. B is given some charge. Under electrostatic and gravitational forces, B may be

Answer»

in equilibrium midway between `A` and `C`
in equilibrium if it is CLOSE to `A` than to `C`
in equilibrium if it is CLOSER to `C` than to `A`
`B` can never be the stable equilibrium.

Answer :B::D
Discussion
2144.

In Young.s experimentusingmonochroniclight ,the fringe pattern shifts by a certain disatnceon the screen when a miea sheet of refractive index1.6and thickness 2 micronis introduced in the path of one of the interfering waves . The mica sheet is then removed and the distancebetween the slits and the screen is doubled .It is foundthat the distance between successive maxima now is the sameat the observed fringe shiftupon the introduction of the mica sheet . The wavelength of light is

Answer»

`5762 A^(0)`
`5825 A^(0)`
`6000 A^(0)`
`6500 A^(0)`

ANSWER :C
Discussion
2145.

Two spheres of Copper of same radii, one hollow and the other solid are charged to the. same potential. Which sphere has more charge?

Answer»

SOLID sphere
Both will HOLD same charge
Hollow sphere
Cannot predict

Answer :B
Discussion
2146.

Abdus Salam, a Pakistan national won Nobel Prize in the field of

Answer»

INELASTIC SCATTERING of light by molecules
Unification of weak and electromagnetic interaction
Superconductivity
Laser technology

Answer :B
Discussion
2147.

The T.V. transmission tower in Delhi has a height of 240 m. The distance up to which the broadcast can be received, (taking the radius of earth to be 6.4 xx 10^(6) m).

Answer»

100 km
60 km
55 km
50 km

Answer :C
Discussion
2148.

A biconvex lens (n = 1.5) has a radius of curvature of magnitude 20 cm. Which one of the following options describe best the image formed of an object of height 2 cm place 30 cm from the lens?

Answer»

real, inverted, height = 1cm
virtual, UPRIGHT, height = 1 cm
virtual, upright, height = 0.5 cm
real, inverted, height = 4 cm

Solution :For the LENS `n=1.5 and |R_(1)|=|R_(2)|=20cm`. As per sign CONVENTION `R_(1)` is `+ve `but `R_(2)` is `-ve`.
`therefore""(1)/(f)=(n-1)((1)/(R_(1))-(1)/(R_(2)))=(1.5-1)((1)/(20)-(1)/((-20)))=(1)/(20) rArr f=+20 cm`
Now `u=-30 cm and H=2cm`
`therefore""(1)/(V)=(1)/(u)+(1)/(f)=(1)/((-30))+(1)/(20)=(1)/(60) rArr v=60cm and h.=(v)/(u)xxh=(60xx2)/((-30))=-4cm`
Thus, theimage is real, inverted and of height 4 cm.
Discussion
2149.

A conducting square loop of side length a is held vertical with the help of a non conducting rod of length L. The rod is made to rotate in horizontal circle about a vertical axis through its end. The loop rotates with the rod while its plane always remains perpendicular to the rod. The resistance of the loop is R and angu- lar speed of the rod is omega. There is a uniform horizontal magnetic field B in the entire space. Find the average rate of heat dissipation in the loop during one rotation.

Answer»


ANSWER :`(a^(4)B^(2)omega^(2))/(2R)`
Discussion
2150.

A concave mirror of radius of curvature 60 cm is placed at the bottom and height of 20 cm. The mirror faces upwards with its axis vertical. Solar light fall normally on the surface of water and image of sun is formed. If amu_omega = 4/3then will the observer in air, the distance of image from the surface of water is

Answer»

30 CM
10 cm
7.5 cm above
7.5 cm below

ANSWER :C
Discussion