+
6. (d) Let O be a point at a distance x from mass m1, where gravitational intensity due to two masses be zero. So
P x
Answer with Explanations
Gm
Total gravitational potential at O
= − − 7. (d) When the vessel was at rest. Equilibrium of block gives
weight =kx+upthrust
or W=kx+F …(i)
As the spring moves downwards with acceleration a(<g) upthrust is reduced.
Now, let x′ be the new compression Then, W−kx′ − ′ =F ma
Therefore, length of spring will increases.
8. (b) If we take the lowest point on the disk as the reference level of gravitational potential energy, the initial total energy is given by
Ei=Ki+Ui=Ui=MgR+Mg R(2 ) = 3MgR
When the object passes beneath the pivot, the potential energy is given by, Uf=MgR+(Mg)(10)=MgR
The final kinetic energy of the system is given by Kf=1Id f + M fR
2 is the moment of inertia of the uniform disk.
The system conservative, so that, Ei=Ef=Kf+Uf
Solving, angular speed of the system ωf g
= 8R
Flow of charge per unit time through any section of the coil = induced current,
i e
Total charge passed through any section between t = 0 s to t = 2s is that subtends an angle ∆θ at the centre as shown in figure. Let the tension in the ring be T.
The forces on this small part ACB tension T by the part of the ring left to A tensinT by the part of the ring right to B.
The tension at A acts along the tangent
at A and the tension at B acts along the tanget at B. As the small part ACB moves in a circle of radius R at a constant speed v.
Its acceleration is towards the centre (along CO) and has a magnitude (∆m v) 2/R
Resolving the forces along the radius CO Tcos 90 Tcos
Substitute the value in Eq. (ii), we get Tension in the ring is T mv
= R
11. (c) Cut this combination of resistances from PQ and let the resistance towards right of PQ be R0 whose value should be such that of connected across AB does not change the entire resistance of the system. Then, the combination is reduced to as
The resistance across PQ=R0+2R Thus, resistance between A and B
= + ×
Initial potential energy across two equal charges, U kQq
i=2 a
and Final potential energy, U kQq
a x a x
13. (c) As per question, it is clear that current through QS should be I 2 in the direction Q to S. Magnetic field induction at a distance r from a long straight conductor carrying current I is
B I
=µ r +
π0 φ1 φ2 4 (sin sin )
When point is at the corner of a long straight conductor, then φ1=90° and φ2= °, so magnetic field induction at M due to0 current through PQ is
B I
acting perpendicular to the plane of paper outwards. The magnetic field induction at M due to current in QR will be zero.
As per question
H I
14. (c) In the given situation, if the magnification of first lens is m, then magnification of second lens in 11m.
Given that, magnification at one half is 2. For L1, object is close so that will be the point with magnification −2, so other half will have m = −1
2 For magnification m v
=u 15. (b) According to conservation of linear momentum,
mvB=(M+m V) ( .0 004×300) ( .= 0 800+0 004. )V
V = 1 493. m/s The frictional force is f=µ(M+m g)
From the work-kinetic energy relationship −fs= ∆KE 1
Then, energy dissipated in the colligion = −
Path difference between the two rays is,
∆ =x CO+PO=(dsecθ+dsec cosθ 2θ)
Phase difference between the two rays is ∆ =φ π (one is reflected, while another is direct)
Therefore, condition for constructive interference should be
∆ =x λ λ
17. (b) According to Heisenberg’s uncertainty principle, we have, So, the proton is non relativistic then,
E=p2/2m, dE pdp
18. (b) When switch S has been closed for a long time, the capacitor gets fully charged and no current flows in arm DG. Then, current in arm C E′ ′ is
Potenial difference across E ′ and D is
′ − = = × =
Potential difference across F and G is VF−VG=I R2 4= 1 ×
40 180 = 4.5 V
∴ VD−Va=(VF−VA) (− VE′ −VD) =4 5 4. − =0 5. V Charge on capacitor C, Q=C V(D−Vc) = ×5 0 5 2 5. = . µC 19. (a) During the growth of voltage in a C-R circuits the voltage
across a capacitor at times t is given by V=Vo[1−e−t CR/ ]
For the given circuit, as per given condition at time t, V =3
4th of voltage applied across capacitor, =3 4Vo Time taken by capacitor
t = ×2 2 5 10. × 6×4 10× −6×0 693 13 86. = . s
Now induced emf = change in flux in unit time
∴ e B vd On suppresing, power saved as
=Pc+Pc× 9 = Pc 64
73 64
∴ % Power saving on suppressed signal
=73 64 × = ×
where, R is the radius of earth At radius of instrument packet, r= 1.1 ,R
Net potential energy of the instrument packet at its maximum altitude,
23. (b) Suppose, the angle of the crown prism needed is A and that of the flint prism is A'.
ω µ µ
The angular dispersion produced by the crown prism is (µv−µr)A=(µ− 1)ωA
Similarly, the angular dispersion produced by the flint prism is (µ′ −1)ω′ ′A
For achromatic combination, the net dispersion should be zero. The deviation in the yellow ray produced by the crown prism is δ=(µ−1 A and by the flint prism is δ) ′ =(µ′ −1 A . The net) ′ deviation produced by the combination is
δ− ′ =δ (µ−1)A−(µ′ −1)A ′
or 1° =0 517. A−0 625. A′ … (ii)
Solving Eqs. (i) and (ii)
A =4.8 and A′ =° 2 4. . Thus, the crown prism should have its° refracting angle 4.8° and that of the flint prism should be 2.4°.
24. (a) In this circuit 20Ω, 30Ω and 60Ω are in parallel. Their effective resistance R1 is 1 1
Here, 24Ω and 8Ω are also in parallel their effective resistance R2 will be Net resistance between X and Y
=3 10+ +6 1 20Ω+ = Current through 8 48
8 6
Ω = = A
Current through 24 48 24 2
Ω = = A
∴ Total current of circuit =6 2+ =8A Therefore potential difference across X and Y
=8 20 160V.× = 25. (c) As, emf across a thermocouple,
E=70Q−
∴ Error in measuring temperature of 200°C of a thermocouple dθ = ±100= ±
50 2 0.
26. (a) Path difference due to slab should be integral multiple of λ
or ∆ =x nλ
At n = 1, the minimum thickness of the glass plate, t = − =
27. (c) The condition for constructive interference in the reflected
system is 2 1
µtcosπ=n+2 λ
where, n = 0 1 2, , , ... and r is angle of refraction,
For normal incidence, i = 0
3 3867Å visible region λ = ×
4 2900=
5 2320Å UV region Therefore refracted light lies in a ultra violet region 28. (c) Current required to operate a 60 W bulb at 120V is
I P
=V= 60 = 120 0 5. A and resistance of bulb
R V
Solving this, we get, value of a capacitor required to operate the lamp, C = 7 7. µF
29. (b) Here kinetic energy of an electron =100eV=100 1 6 10× . × −9J value this lost when electron moves through a distance (d) towards the negative plate.
KE = work done = × =F S qE×S Distance travelled by an electron,
d = × × × ×
30. (d) In common emitter configuration, the current gain is
A h
hoe= output admittance
∆
Current gain
Ai= −