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Jee Main Advanced Physics Chapter Wise Mock Test
Quiz 3
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Q.1
A heated object emits radiation which has maximum intensity at frequency m.Now if the temperature of the body is doubled,then the maximum intensity of radiation happens at the frequency
0%
4 m
0%
2 m
0%
1 m
0%
.5 m
Explanation
Solution:
Wien's law for the FREQUENCY spectrum runs the opposite way to the wavelength version: peak frequency is directly proportional to temperature (f
max
∝T), not inversely. So doubling T simply doubles the frequency at which the peak occurs.
Q.2
An electron of mass me initially at rest moves through a certain distance in a uniform electric field in time tA proton of mass mp also initially at rest ,takes time t2 to move through an equal distance in this uniform electric field.Neglect any effect of gravity. Which of the following equation is correct in this case
0%
met22 -mpt12 =0
0%
met12 -mpt22 =0
0%
met2 -mpt1 =0
0%
None of the above
Explanation
Solution:
Both start from rest and cover the same distance d under the same field E, so d=½(qE/m)t² for each, using the same charge magnitude q but their own mass and time.
Equating the two: (1/m
e
)t
1
² = (1/m
p
)t
2
² ⇒ m
p
t
1
² = m
e
t
2
² ⇒ m
e
t
2
² - m
p
t
1
² = 0.
Q.3
In the visible region,the dispersive powers for crown and flint glass prism are ω1 and ω2 , and the mean angular deviation are d1 and d2 respectively. When the two prism are combined ,the condition of zero dispersion by the combination is
0%
$(\omega _1 d_1)^2 + (\omega _2 d_2)^2 =0$
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$(\omega _1 d_2) + (\omega _2 d_1) =0$
0%
$(\omega _1 d_1) + (\omega _2 d_2) =0$
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$\sqrt {\omega _1 d_1}+ \sqrt {\omega _2 d_2} =0$
Explanation
Solution:
For two prisms combined so their dispersions cancel (an achromatic combination), the individual dispersions -- each prism's dispersive power times its own mean deviation -- must sum to zero, with the prisms oriented so one's contribution offsets the other's: ω
1
d
1
+ ω
2
d
2
= 0.
Q.4
A train sounds its whistle as it approaches and leaves a railroad crossing. An observer at the crossing measures a frequency of 219 hz as the train approaches and a frequency of 184 hz as the train leaves.The speed of the sound is 340 m/s. Find the speed of the train and frequency of its whistle
0%
26.5 m/s, 200 hz
0%
29.5 m/s, 150 hz
0%
29.5 m/s, 200 hz
0%
29.5 m/s, 250 hz
Explanation
Solution:
f
1
/f
2
= (v+v
s
)/(v-v
s
) ⇒ 219/184 = (340+v
s
)/(340-v
s
).
219(340-v
s
) = 184(340+v
s
) ⇒ 340(219-184) = 403v
s
⇒ v
s
= 11900/403 ≈ 29.5 m/s.
From f
1
=f
0
v/(v-v
s
): f
0
= 219×(340-29.5)/340 ≈ 200 Hz.
Q.5
A larger wooden wheel of radius R and moment of inertia 'I' is mounted on an axle so as to rotate freely. A bullet of mass m and speed v is shot tangential to the wheel and strikes its edge ,lodging in the rim. If the wheel were originally at rest. What will be angular velocity after the collision
0%
$ \frac {mvr}{mR^2 + I}$
0%
$ \frac {mR^2}{mvr + I}$
0%
$ \frac {mR^2}{mvr +2 I}$
0%
None of the above
Explanation
Solution:
Angular momentum is conserved about the wheel's axle: the bullet's initial angular momentum, mvR (tangential hit at radius R), equals the final angular momentum of the whole system (wheel plus the now-embedded bullet):
mvR = (I+mR²)ω ⇒ ω = mvR/(mR²+I).
Q.6
Two blocks of mass m1= m2=3Kg are connected by a light string.They are placed on a rough horizontal surface. A force of magnitude 20 N is applied in the horizontal direction on a block,So that both the masses move with acceleration .5 m/s2.Let T be the tension in the string.The value of T/10 is
0%
1
0%
2
0%
4
0%
3
Explanation
Solution:
For the whole two-block system (6 kg total): F - f
total
= (6)(0.5) = 3N, so f
total
= 20-3 = 17N, giving μ = 17/(6×10) ≈ 0.283.
For the trailing block alone (only pulled by tension T, held back by its own share of friction, f
2
=μ(3)(10)=8.5N): T - f
2
= (3)(0.5) = 1.5N, so T = 1.5+8.5 = 10N, and T/10 = 1.
Q.7
A body executing linear SHM has a velocity of 3 cm/s when its displacement is 4 cm and velocity of 4 cm/s when its displacement is 3 cm. Find the amplitude of oscillation
0%
2 cm
0%
3 cm
0%
4 cm
0%
5 cm
Explanation
Solution:
For SHM, v²=ω²(A²-x²). The two data points give 9=ω²(A²-16) and 16=ω²(A²-9).
Dividing: 16/9 = (A²-9)/(A²-16) ⇒ 16(A²-16)=9(A²-9) ⇒ 7A²=175 ⇒ A²=25 ⇒ A=5 cm.
Q.8
If earth were to suddenly contract to half of its present size with out change in any mass.What will be the duration of the day?
0%
4
0%
8
0%
10
0%
6
Explanation
Solution:
With no external torque, angular momentum Iω is conserved. For a uniform sphere, I=(2/5)MR², so halving R quarters I -- meaning ω must quadruple to keep Iω the same.
Since the day length T=2π/ω, quadrupling ω quarters the day: 24/4 = 6 hours.
Q.9
Ionization potential of hydrogen atom is 13.6 V. Hydrogen in the ground state is excited by monochromatic radiation of photons of energy 12.1 eV. The number of spectral lines emitted by the hydrogen atom according to Bohr's theory will be
0%
1
0%
2
0%
3
0%
4
Explanation
Solution:
Absorbing 12.1 eV raises the electron from -13.6 eV to -13.6+12.1=-1.5 eV. Since E
n
=-13.6/n², solving -13.6/n²=-1.5 gives n²≈9, so n=3.
From n=3, every possible downward transition can occur (3→2, 3→1, 2→1) -- 3 spectral lines, matching the general count of C(n,2) possible pairs.
Q.10
A spherical surface of radius of curvature R separates air(refractive index 1.0) from glass (refractive index 1.5).The center of curvature is in the glass .A point object S1 placed in air is found to have a real image S2 in the glass.The line S1S2 cuts the surface at point O such that S1O = OSThe value of S1O/R is
0%
4
0%
1
0%
3
0%
5
Explanation
Solution:
Using n
2
/v - n
1
/u = (n
2
-n
1
)/R with the object in air (u=-d) and a real image in glass (v=+d), where d=S
1
O=OS
2
:
1.5/d - 1/(-d) = (1.5-1)/R ⇒ 1.5/d+1/d = 0.5/R ⇒ 2.5/d=0.5/R ⇒ d = 5R.
So S
1
O/R = 5.
Q.11
Centrifugal force is an inertial force when considered by
0%
An observer who is moving with the particle that experience the force
0%
an outside observer
0%
An observer at the center of the circular motion
0%
None of the above
Explanation
Solution:
Centrifugal force isn't a real, physically-caused force -- it's a bookkeeping trick needed only when you insist on analysing motion from inside the rotating (non-inertial) frame of the object itself. An outside observer, watching from a genuinely inertial frame, only ever needs the real centripetal force -- no outward counterpart is required.
Q.12
Choose the correct statement
0%
Two important differences between viscous forces and frictional forces are that viscous force depends upon the temperature and also velocity profile for the liquid where as frictional forces is usually independent of such items
0%
The terminal velocity of a spherical body falling in a fluid depends on the diameter and density of the body as well the density and viscosity of the fluidcorrect
0%
M drops of a fluid ,each with surface energy k,join to form a single drop,the amount of energy released in the process is K(M - M^{2/3})correct
0%
Tiny insects can float and walk on the surface of water due to both buoyancy and surface tensioncorrect
Explanation
Solution:
Viscous drag depends on temperature and the fluid's velocity profile, unlike ordinary dry friction -- true. Stokes' law makes terminal velocity depend on the body's size and density along with the fluid's density and viscosity -- true. For M identical drops (each surface energy k) coalescing into one, conserving volume while shrinking total surface area releases energy of the form K(M-M^(2/3)) -- the standard result for this classic problem -- true. And insects resting on water are supported mainly by surface tension, with a small buoyant contribution too -- also generally considered true.
Q.13
A pendulum is constructed from two identical uniform rods a and b each of length L and mass M,connected at right angles to form a T by joining the center of rod a to one end of the rod b. The T pendulum is then suspended from free end of rod b and pendulum swings in the plane of T The moment of inertia of T about the axis of rotation
0%
$ \frac {11ML^2}{12}$
0%
$ \frac {7ML^2}{12}$
0%
$ \frac {17ML^2}{18}$
0%
$ \frac {17ML^2}{12}$
Explanation
Solution:
Rod b, pivoted at its own end: I
b
=(1/3)ML².
Rod a, joined at its own centre to the far end of rod b (a distance L from the pivot): by the parallel axis theorem, I
a
= (1/12)ML² + ML² = (13/12)ML².
Total: I = (1/3)ML² + (13/12)ML² = (4/12)ML²+(13/12)ML² = (17/12)ML².
Q.14
The potential energy change of the T when it is rotated by angle $\theta$ from vertical position
0%
$\frac {3Mg}{L}(1-\cos \theta)$
0%
$\frac {3Mg}{L}(1-\sin \theta)$
0%
$\frac {4Mg}{L}(1-\sin \theta)$
0%
$\frac {2Mg}{L}(1-\sin \theta)$
Explanation
Solution:
The whole T's centre of mass sits at a distance d
cm
=[M(L/2)+M(L)]/(2M) = (3/4)L from the pivot (rod b's own CM contributes L/2, rod a's CM sits right at the junction, a full L away).
Swinging by angle θ raises that centre of mass by d
cm
(1-cosθ), so ΔU = (2M)g×(3L/4)(1-cosθ) = (3MgL/2)(1-cosθ).
Q.15
For small angle with vertical position,The T makes SHM,the angular frequency
0%
$ \sqrt { \frac {2g}{17L}}$
0%
$ \sqrt { \frac {17g}{19L}}$
0%
$ \sqrt { \frac {12g}{11L}}$
0%
$ \sqrt { \frac {18g}{17L}}$
Explanation
Solution:
For a physical pendulum, ω² = (total weight × distance to CM) ÷ (moment of inertia about the pivot) = [2Mg×(3L/4)] ÷ [17ML²/12] = (3MgL/2)×(12/(17ML²)) = 18g/(17L).
ω = √(18g/(17L)).
Q.16
A vertical cylinder of cross-sectional area S contains one mole of ideal monoatomic gas under a piston of Mass M. At t=0, an energy source which transfer Q per unit time to gas is switched on under the piston. The gas pressure remains constant and equal to PAnd the gas is thermally insulated. What will be the established velocity of the piston?
0%
$v=\frac {2}{5} \frac { Q}{(P_0 A+Mg)}$
0%
$v=\frac {52}{52} \frac { Q}{(P_0 A+Mg)}$
0%
$v= \frac { Q}{(P_0 A+Mg)}$
0%
None of the above
Explanation
Solution:
At constant pressure, dQ=nC
p
dT (C
p
=(5/2)R for a monatomic gas), and since PV=nRT with P fixed, V∝T, so dV/dt=(nR/P)(dT/dt).
Piston speed v=(1/S)(dV/dt), and dT/dt=Q/(nC
p
)=2Q/(5nR) (from the given heat rate Q):
v = (1/S)(nR/P)×2Q/(5nR) = 2Q/(5PS).
Since the gas pressure P is set by supporting both the atmosphere and the piston's weight, P=P
0
+Mg/S, substituting gives v = (2/5)Q/(P
0
S+Mg).
Q.17
Link Comprehension Type Use this to solve the Question 2 and 3 An object moves along a circular path of radius R with deceleration. At any moment, the magnitudes of tangential and normal acceleration are same. The object starts with velocity v0 Find the velocity v of the object as function of distance s covered by it
0%
$v=v_0 e^(-R/s)$
0%
$v=v_0 e^(-2s/R)$
0%
$v=v_0 e^(-s/2R)$
0%
$v=v_0 e^(-s/R)$
Explanation
Solution:
a
t
=v(dv/ds) and a
n
=v²/R. Setting their magnitudes equal, with the object decelerating (dv/ds<0): v(dv/ds) = -v²/R.
Separating variables: dv/v = -ds/R. Integrating from v
0
at s=0: ln(v/v
0
) = -s/R, so v = v
0
e
-s/R
.
Q.18
The total acceleration as a function of distance s covered
0%
$a=\frac {\sqrt {2} v_0^2 e^(-2s/R)}{R}$
0%
$a=\frac {v_0^2 e^(-2s/R)}{R}$
0%
$
a=\frac {\sqrt {3} v_0^2 e^(-2s/R)}{R}
0%
None of these
Explanation
Solution:
With a
t
and a
n
always equal in magnitude, the total acceleration is √(a
t
²+a
n
²) = √2×a
n
= √2×v²/R.
Using v²=v
0
²e
-2s/R
from the previous part: a = √2·v
0
²e
-2s/R
/R.
Q.19
A Mass M with small electric charge q slides on a smooth inclined plane of angle θ to the horizontal. A magnetic Field is directed perpendicular to the section of the plane. Find the acceleration of the mass when the velocity is v
0%
$\frac (Bvq-gsin \theta)}{M}$
0%
$\frac (Bvq+gsin \theta)}{M}$
0%
$\frac (Bvq-2gsin \theta)}{M}$
0%
$g \sin \theta $
Explanation
Solution:
Two effects act along the slope: gravity's component Mg sinθ drives the charge down the incline, while the magnetic force from its own motion through the perpendicular field B grows with speed (∝qvB) and increasingly opposes that acceleration as v builds up -- giving a net acceleration of (Bvq-Mg sinθ)/M that shrinks as the charge speeds up, the same qualitative behaviour as any velocity-dependent braking effect approaching a terminal speed.
Q.20
Link Type Comprehension Use this to solve the Question 5, 6 and 7 Two resistor 4.4 and 8.8 ohm and two capacitors .48 and .24 µF are arranged as shown in below figure. A potential difference 24 V is applied across the combination as shown below. The negative terminal of the battery is assumed to be zero potential Find the potential of point of a and b when switch S is open
0%
4 V,12 V
0%
2 V ,14 V
0%
8 V ,16 V
0%
122V,778V
Explanation
Solution:
With S open, current only flows through the two resistors (in series): I=24/(8.8+4.4)=24/13.2≈1.818A, giving V
a
=24-I(8.8)=8V.
The two capacitors (also in series, isolated except through this same 24V) share a common charge at steady state: C
eq
=(0.48×0.24)/(0.48+0.24)=0.16μF, so Q=0.16×24=3.84μC, and V
b
(the node between them) = 24 - Q/0.48 = 24-8 = 16V.
So V
a
=8V, V
b
=16V.
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