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Question 1: A current carrying wire heats a metal rod. The wire provides a constant power P to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes with time (t) as T(t) = T0 (1 + βt1/4) where β is a constant with appropriate dimension of temperature. The heat capacity of metal is:
Question 2: In a capillary tube of radius 0.2 mm the water rises up to height of 7.5 cm with angle of contact equal to zero. If another capillary with same radius but of different material dipped in the same liquid. The height of water raised in capillary will be, if angle of contact becomes 60°.
T = Rhρg/2cosθ
h/cosθ = constant
7.5/1 = h’/(1/2)
⇒ h’ = 3.75 cm
Question 3: A sample of 19K40 disintegrates into two nuclei Ca & Ar with decay constant λCa = 4.5 x 10-10 S-1 and λAr = 0.5 x 10-10 S-1 respectively. The time after which 99% of 19K40 gets decayed is:
λ = λ
(1/100) N0 = N0 e-λt
ln(1/100) = -( λ1 + λ2)t + ln 100 = +( λ1 + λ2)t2
⇒ (2.303x2)/(5x10-10) = t
Or t = 9.2 x 109 sec
Question 4: Consider a spherical gaseous cloud of mass density ρ(r) in a free space where r is the radial distance from its centre. The gaseous cloud is made of particles of equal mass m moving in circular orbits about their common centre with the same kinetic energy K. The force acting on the particles is their mutual gravitational force. If ρ(r) is constant with time. The particle number density n(r) = ρ(r)/m is:
(g = universal gravitational constant)
GMm/r = mv2/r = 2/r x ½ x mv2 = 2k/r
M = 2kr/Gm
4πr2 dr ρ = 2k dr/Gm
Or ρ = k/2πGmr2
Question 5: A thin spherical insulating shell of radius R caries a uniformly distributed charge such that the potential at its surface is V0. A hole with small area α4πR2 (α <<<1) is made in the shell without effecting the rest of the shell. Which one of the following is correct.
dq = Q/4πR2 - dA = Qα
Vc = KQ/R - KαQ/R = V0(1- α)
VB = KQ/R - KαQ/[R/2] = V0(1 - 2α)
VA = KQ/(2R)2 - KαQ/R2 = KQ/4R2 - αV0/R
Reduced by αV0/R
So E@C = kαQ/R2 = αV0/R
Increased by 2V0/R
Question 6: A charged shell of radius R carries a total charge Q. Given ϕ as the flux of electric field through a closed cylindrical surface of height h, radius r & with its center same as that of the shell. Here center of cylinder is a point on the axis of the cylinder which is equidistant from its top & bottom surfaces. Which of the following are correct.
Case 1: If h > 2R
Q = Q/E0
Case 2: h = 8R/5
r = 3R/5
Using Gauss law concept ACD are correct
Question 7: Which statements is/are corect:
When S1 is closed then charge on capacitor is zero.
While replacing all capacitors with wire, we have
i = t/[70+100+30] = 5/200 = 25 mA
When circuit is in steady state, q/10 + q/80 + q/80 - 5 = 0
or q = 40 μC
Therefore, voltage across C1 = 40/10 = 4 volt
Just after closing of S2, charge on capacitor remain same, KVL
or 30x + 70y = 6
Again, -40/80 + 5 + (x-y)30 - 40/80 + (x - y) 100 - 10 + 30x = 0
⇒ 160x - 130y - 6 = 0
Solving both the equations, we have x = 0.05 amp and y = 96/1510
Option (b) and (c) are correct.
Question 8: A galvanometer of resistance 10 ohm and maximum current of 2μA is converted into voltmeter of range 100 mV and when converted into ammeter then range is 1mA. When these voltmeter and ammeter are connected by a (ideal) battery in series with a resistance of R = 1000 Ω, then
V = 100 x 10-3 V
V = Ig(Rg + R)
= (Rg + R)Rγ
RV = 5 x 104
Question 9: Conducting wire of parabolic shape, initially y = x2 is moving with velocity
B, C is correct
D is also correct because projection of wire on y axis is same.
Question 10: If in a hypothetical system if the angular momentum and mass are dimensionless. Then which of the following is true.
[M] = [M0L0T0]
[J] = [ML2T-1]
[ML2T-1] = [M0L0T0]
⇒ [L2] = [T]
[P] = [MLT-2 . LT-1] = [ML2T-3] = [L2L-6] = [L-4]
[E] = [MLT-2 . L] = [L2L-4] = [L-2]
[F] = [LL-4] = [L-3]
Options A,B,C are correct.
Question 11: V – T diagram for n mol monoatomic gas is given below:
Corresponding PV entraps
(d) There are no adiabatic process is involved.
Only Options (a) and (b) are correct.
Question 12: Apparent depth for point object x in all three cases are H1, H2 & H3 respectively when seen from below given H = 30 cm, n = 1.5 & R = 3m, then
⇒ 1/n = d1/(3Q)
⇒ d1 = (30/3) x 2 = 20 cm
H2 = 20.684
V3 = 19.354
Consider the following nuclear fission reaction
In this fission reaction. Kinetic energy of α-particle emitted is 4.44 MeV. Find the energy emitted as γ – radiation in keV in this reaction.
m(88Ra226) = 226.005 amu
m(2He4) = 4.000 amu
Δm = 0.005 amu
Question 14: N dielectrics are introduced in series in a capacitor of thickness D. Each dielectric have width d = D/N &
dielectric constant of mth dielectric is given by Km = K(1 + m/N) : [N >> 103, Area of plates = A]
Net capacitance is given by [KЄoA]/[αDln2]. Find value of α.
x/m = Δ/N
Integrating we get
Ceq = [KЄoA]/[Dln2]
Therefore, α = 1
Question 15: If at angle θ the light takes maximum time to travel in optical fiber. Then the maximum time is x (10-8), calculate x.
1.5 sin θC = 1.44 sin 900
sin θC = 24/25
and sin θC = x/d
⇒ d = 25x/24
t = L/(C/n2) = 5 x 108 s
Putting: C = 3 x 108, n2 = 1.5 and L = 10
Question 16: The source S1 is at rest. The observer and the source S2 are moving towards S1 as shown in figure. The roof beats observed by the observer if both sources have frequency 120 Hz and speed of sound 330 m/s in is
Question 17: A weight of 100 N is suspended by two wires made by steel and copper as shown in figure length of steel wire is 1 m and copper wire is √3 m. Find ratio of change in length of copper wire (Δ l0) to change in length of steel wire (Δ ls). Given Young’s modulus: Ysteel = 2 × 1011 N/m2, Ycopper = 1 × 1011 N/m2.
Question 18: An optical bench, to measure the focal length of lens, is 1.5 m long and on the bench marks are with spacing 1/4 cm. Now a lens is placed at 75 cm and pin type object is placed at 45 cm marks on the bench. If its image is formed at 135 cm find maximum possible error in calculation of focal length.
V = 30 cm and dv = 0.5 cm
V = 60 cm and dv = 0.5 cm
Now, using lens formula,
1/v - 1/u = 1/f
1/60 + 1/30 = 1/f
⇒ f = 20 cm
= (20/2)[1/602 + 1/302]
Multiply by 100 both the sides, we get
df/f = 1.38 and 1.39