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Jee Main Advanced Physics Chapter Wise Mock Test
Quiz 1
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Q.1
A big drop of water is broken into large number of small drops? The surface energy would
0%
Remains unchanged
0%
will increase
0%
will decrease
0%
Not enough information
Explanation
Solution:
Splitting one big drop into many small drops keeps the total volume the same but spreads it across far more total surface area (many small spheres have more combined surface than one large sphere of the same volume). Since surface energy = surface tension × surface area, more surface area means more surface energy -- so it increases.
Q.2
A U tube containing a liquid is accelerated horizontally with constant acceleration a .The separation between the limb's is L. The Difference in the height of the liquid in the two arms would be
0%
$L \sqrt { \frac {a}{g}}$
0%
$\frac {aL}{g}$
0%
$ \frac {L}{2}$
0%
$\frac {a^2L}{g^2}$
Explanation
Solution:
In the accelerating frame of the tube, the liquid experiences an effective backward pseudo-force, tilting its free surface just like an effective extra "gravity" component a acting horizontally alongside the real g acting vertically.
The free surface settles at an angle θ where tanθ=a/g, and over the horizontal separation L between the two limbs, the height difference is Δh = L tanθ = aL/g.
Q.3
A block of wood has a mass 25 g. When a 5 g metal piece with a volume 2 cm3 is attached to the bottom of the block,the wood barely floats in water what is the volume of the V of the wood
0%
$20 cm^3$
0%
$38 cm^3$
0%
$28 cm^3$
0%
None of these
Explanation
Solution:
"Barely floats" means the whole combination (wood + metal) is just on the verge of being fully submerged, so its total weight equals the weight of water displaced by its FULL combined volume (V for the wood plus 2 cm³ for the metal):
(25+5) g = 1 g/cm³ × (V+2) cm³ ⇒ 30 = V+2 ⇒ V = 28 cm³.
Q.4
A solid sphere of radius R,made up of a material of bulk modulus K is surrounded by a liquid in a cylindrical container. A mass-less piston of area A floats on the surface of the liquid.When a mass M is placed on the piston to compress the liquid,the fractional change in the radius of the sphere is
0%
$ \frac {Mg}{2AK}$
0%
$ \frac {3Mg}{AK}$
0%
$ \frac {Mg}{3AK}$
0%
$ \frac {Mg}{AK}$
Explanation
Solution:
The mass M on the piston adds extra pressure ΔP = Mg/A throughout the liquid (Pascal's principle), including at the sphere's surface.
Bulk modulus: K = -ΔP/(ΔV/V), so ΔV/V = -Mg/(AK) in magnitude.
For a sphere, V=⁴ₛ₃πR³, so V is proportional to R³, meaning a small fractional volume change relates to a fractional radius change by ΔV/V = 3ΔR/R.
So ΔR/R = ⅓(ΔV/V) = ⅓ × Mg/(AK) = Mg/(3AK).
Q.5
A small hole is there near the bottom of the water filled container. The speed of the water ejected depends on
0%
Density of the liquid
0%
acceleration due to gravity
0%
height of the liquid above the hole
0%
All of the above
Explanation
Solution:
By Bernoulli's equation between the liquid's top surface and the hole, ρgh = ½ρv² -- the density ρ appears on both sides and cancels out completely, leaving v=√(2gh): the efflux speed depends only on g and the height of liquid above the hole, not on what liquid it is.
Q.6
if the hollow bob of a simple pendulum be filled with mercury that drains out slowly,its time period
0%
increases continuously
0%
decreases continuously
0%
remains same
0%
first increases and then decreases
Explanation
Solution:
As mercury drains out through the hole at the bottom, the mercury that remains settles under gravity into a shrinking puddle at the BOTTOM of the hollow bob (not the top) -- so early in the draining, the mercury's own centre of mass moves DOWN, away from the pivot, increasing the effective pendulum length and hence increasing the period.
But as draining continues and the remaining mercury's mass becomes small compared to the (fixed) shell, the combined centre of mass has to swing back toward the shell's own centre of mass (at the sphere's geometric centre) once the bob is nearly empty -- pulling the effective length, and the period, back down again. So the period first increases, then decreases.
Q.7
Water leaves a faucet with a downward velocity of 3 m/s. As the water falls below the faucet,it accelerates with acceleration g. The cross-section area of the water stream leaving the faucet is 1.0 cmWhat is the cross-sectional area of the stream .5 m below the faucet?
0%
.50 cm2
0%
.9 cm2
0%
.1 cm2
0%
.69 cm2
Explanation
Solution:
Using v²=v
0
²+2gh with v
0
=3 m/s and h=0.5 m: v² = 9 + 2(9.8)(0.5) = 18.8, so v ≈ 4.34 m/s.
By continuity, A
1
v
1
=A
2
v
2
, so A
2
= A
1
v
1
/v
2
= (1.0×3)/4.34 ≈ 0.69 cm².
Q.8
Machine parts are jammed in winter due to
0%
Increase in surface tension of lubricant
0%
Decrease in viscosity of lubricant
0%
Decrease in surface tension of lubricant
0%
increase in viscosity of lubricant
Explanation
Solution:
A liquid's viscosity rises as it gets colder -- this is exactly why engine oils are given winter viscosity grades, since cold oil flows far more sluggishly. That thicker, harder-to-flow lubricant is what makes machine parts stiffen up and jam in cold winter weather.
Q.9
The dimensions of viscosity in terms of M,L,T is ?
0%
$MLT^{-1}$
0%
$M^{-1}LT^{-1}$
0%
$ML^{-1}T^{-2}$
0%
$ML^{-1}T^{-1}$
Explanation
Solution:
Viscosity is defined through F = ηA(dv/dx), so η = F ÷ [A·(dv/dx)].
Dimensionally: [MLT⁻²] ÷ ([L²]×[T⁻¹]) = [MLT⁻²] ÷ [L²T⁻¹] = [ML⁻¹T⁻¹].
Q.10
A man is sitting in a boat which is floating on a pond. The man drinks some water from pond. What happens to the water level in the pond? The water level will
0%
Rises
0%
falls
0%
remains same
0%
Not enough information
Explanation
Solution:
The floating boat+man system always displaces exactly its own weight in water (Archimedes' principle). Drinking water removes some water from the pond, but that same water is now inside the man's body, adding exactly that much extra weight to the boat+man system -- which then must displace exactly that much MORE water to stay afloat. The water taken out of the pond and the extra water displaced by the heavier boat are equal, so they cancel and the pond's level doesn't change.
Q.11
A body floats in water with 40% of its volume outside water.When the same body floats in some liquid,60% of its volume remains outside the liquid. The relative density of the liquid is
0%
1.5
0%
1.2
0%
.6
0%
None of these
Explanation
Solution:
In water: 40% outside means 60% submerged, so by Archimedes' principle ρ
body
= 0.6ρ
water
(the fraction submerged equals the density ratio).
In the other liquid: 60% outside means 40% submerged, so ρ
body
= 0.4ρ
liquid
.
Combining: 0.4ρ
liquid
= 0.6ρ
water
, so ρ
liquid
= 1.5ρ
water
-- a relative density of 1.5.
Q.12
A closed compartment containing gas is moving with some acceleration in horizontal direction. Then the pressure in the compartment is? Neglect the effect of gravity
0%
lower in the front side
0%
same everywhere
0%
lower in the rear side
0%
Not enough information
Explanation
Solution:
In the compartment's own accelerating frame, gas experiences an effective pseudo-force pointing opposite to the acceleration (i.e. toward the rear), the same way passengers feel pushed back into their seats. This piles gas up toward the rear, raising pressure there, and leaves the front with correspondingly lower pressure -- the same reason a helium balloon in an accelerating car drifts toward the front, toward the region of lower pressure.
Q.13
A object of relative density 10 is released from rest on the surface of a lake.if the viscous effect are ignored ,the object sinks in the water with an acceleration ?
0%
10g
0%
9g/10
0%
11g/10
0%
None of these
Explanation
Solution:
As the object sinks, two forces act: its weight (down) and buoyancy (up). Net acceleration = g × (1 - ρ
water
/ρ
object
).
With relative density 10 (ρ
object
=10ρ
water
): a = g(1 - 1/10) = 9g/10.
Q.14
A horizontal pipeline carries water in a stream line flow? At point A along the pipe,the cross-sectional area is 10 cm2,the water velocity is 1 m/s and pressure is 2000 Pa.What is the pressure at point B where cross-sectional area is 5.0 cm2
0%
500 Pa
0%
400 Pa
0%
300 Pa
0%
None of the above
Explanation
Solution:
Continuity: A
1
v
1
=A
2
v
2
⇒ v
2
= (10×1)/5 = 2 m/s.
Bernoulli's equation for a horizontal pipe (no height change): P
1
+½ρv
1
² = P
2
+½ρv
2
².
P
2
= 2000 + ½(1000)(1²-2²) = 2000 + 500(-3) = 2000-1500 = 500 Pa.
Q.15
Water rises to a height of 13.6 cm in a capillary tube dipped in water.When the same tube is dipped in mercury ,it is depressed by $3 \sqrt {2}$ cm. The angle of contact in water is 0°. The angle of contact in mercury =135° Given : Relative density of mercury =13.6 Find out the ratio of the surface tensions of mercury and water
0%
6
0%
5
0%
11
0%
None of these
Explanation
Solution:
Capillary rise/depression: h = 2Tcosθ/(ρgr).
Water: 13.6 = 2T
w
cos(0°)/(ρ
w
gr) = 2T
w
/(ρ
w
gr).
Mercury: 3√2 = 2T
Hg
|cos(135°)|/(ρ
Hg
gr) = T
Hg
√2/(ρ
Hg
gr).
Dividing the mercury equation by the water equation (and using ρ
Hg
=13.6ρ
w
):
(3√2)/13.6 = [T
Hg
√2/(13.6ρ
w
gr)] ÷ [2T
w
/(ρ
w
gr)] = T
Hg
/(2×13.6×T
w
/√2)
Solving this out gives T
Hg
/T
w
= 6.
0 h : 0 m : 1 s
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