As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background.
Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .
Q1. Typical silt(hard mud) particle of radius 20μm is on the top of lake water, its density is 2000 kg m(-3) and the viscosity of lake water is 1.0 mPa, density is 1000 kg m(-3). If the lake is still(has no internal fluid motion). The terminal speed with which the particle hits the bottom of the lake is ………. mms(-1)
Solution
v=2/9 (r2 (ρ-σ)g)/ɳ =2/9×((20×10(-6) )2] (2000-1000)×9.8)/(1.0×10(-3) ) =8.7×10(-4) ms(-1)=0.87mm s(-1)
v=2/9 (r2 (ρ-σ)g)/ɳ =2/9×((20×10(-6) )2] (2000-1000)×9.8)/(1.0×10(-3) ) =8.7×10(-4) ms(-1)=0.87mm s(-1)
Q2.Water rises in a capillary tube to a height h. Choose false statement regarding capillary rise from the following.
Solution
The height (h) to which water rises in a capillary tube is given by h=(2T cosθ)/rρg where θ is angle of contact , r the radius, ρ the density and g acceleration due to gravity. When lift moves down with constant acceleration, height is less than h, because effective value of acceleration due to gravity increases hence h decreases.
The height (h) to which water rises in a capillary tube is given by h=(2T cosθ)/rρg where θ is angle of contact , r the radius, ρ the density and g acceleration due to gravity. When lift moves down with constant acceleration, height is less than h, because effective value of acceleration due to gravity increases hence h decreases.
Q3. A horizontal pipe line carries water in streamline flow. At a point where the cross-sectional area is 10 cm2 the water velocity is 1 ms(-1) and pressure is 2000 Pa. The pressure of water at another point where the cross-sectional area is 5 cm2, is
Solution
Since cross-sectional area is halved, therefore, velocity is doubled. Now, p1=2000Pa,v1=1ms(-1) p2=?,v2=2 ms(-1) Again p2+1/2×1000×2×2=2000+1/2×1000×1×1 or p2=2000+500(1-4)=500Pa
Since cross-sectional area is halved, therefore, velocity is doubled. Now, p1=2000Pa,v1=1ms(-1) p2=?,v2=2 ms(-1) Again p2+1/2×1000×2×2=2000+1/2×1000×1×1 or p2=2000+500(1-4)=500Pa
Q4. Radius of an air bubble at the bottom of the lake is r and it becomes 2rwhen the air bubble rises to the top surface of the lake. If ρ cm of water be the atmospheric pressure, then the depth of lake is
Solution
Let depth of lake is x cm. ∴ p1 V1=p2 V2 (pdg+xdg)(4/3 πr3 )=pdg[4/3 π(2r)3 ] (p+x) r3=p(8r3 ) x=8p-p x=7p
Let depth of lake is x cm. ∴ p1 V1=p2 V2 (pdg+xdg)(4/3 πr3 )=pdg[4/3 π(2r)3 ] (p+x) r3=p(8r3 ) x=8p-p x=7p
Q5.Water is flowing through a horizontal pipe of varying cross-section. If the pressure of water equals 2 cm of mercury, where the velocity of the flow is 32 cm s(-1), what is the pressure at another point, where the velocity of flow is 65 cm s(-1) ?
Solution
Here, p_1=2 cm of Hg =2×13.6×980 =2.666×104 dyne cm(-2) v1=32 cms(-1), v2=65cms(-1) For a horizontal pipe, according to Bernoulli’s theorem, p1/ρ+1/2 v12=p2/ρ+1/2 v22 or p2=p1+1/2 ρ(v12-v22 ) or p2=2.666×104+1/2×[(32)2-(652)] =2.666×104-0.16×104 = (2.506×104)/(13.6×980)=1.88 cm of Hg
Here, p_1=2 cm of Hg =2×13.6×980 =2.666×104 dyne cm(-2) v1=32 cms(-1), v2=65cms(-1) For a horizontal pipe, according to Bernoulli’s theorem, p1/ρ+1/2 v12=p2/ρ+1/2 v22 or p2=p1+1/2 ρ(v12-v22 ) or p2=2.666×104+1/2×[(32)2-(652)] =2.666×104-0.16×104 = (2.506×104)/(13.6×980)=1.88 cm of Hg
Q6. A vessel whose bottom has round hole with diameter of 1 mm is filled with water. Assuming that surface tension acts only at hole, then the maximum height to which the water can be filled in vessel without leakage is(surface tension of water =7.5×〖10〗(-2) Nm(-1) and g=10ms(-2))
Solution
Water will not leak out from the hole if the weight of water in the water column is supported by the force due to surface tension. Using the relation, h=2T/rρg =(2×7.5×10(-2)/(0.5×10(-3)×103×10) =3×10(-2) m=3 cm
Water will not leak out from the hole if the weight of water in the water column is supported by the force due to surface tension. Using the relation, h=2T/rρg =(2×7.5×10(-2)/(0.5×10(-3)×103×10) =3×10(-2) m=3 cm
Q7.In a capillary rise experiment, the water level rises to a height of 5 cm. If the same capillary tube is placed in water such that only 3 cm of the tube projects outside the water level, then
Solution
water will rise to a level less than 3 cm
water will rise to a level less than 3 cm
Q8.Two stretched membranes of area 2 cm2 and 3 cm2 are placed in a liquid at the same depth. The ratio of pressure on them is
Solution
Pressure is independent of area of cross-section
Pressure is independent of area of cross-section
Q9.A balloon of volume 1500 m3 and weighing 1650 kg with all its equipment is filled with He (density0.2 kg m(-3). If the density of air be 1.3 kgm(-3), the pull on the rope tied to the balloon will be
Solution
Pull on the rope = effective weight =[1650+(1500×0.2)-1500×1.3] kgf =1650+300-1950 =0
Pull on the rope = effective weight =[1650+(1500×0.2)-1500×1.3] kgf =1650+300-1950 =0
Q10. A steel ball is dropped in oil then,
Solution
According to Stokes, when a spherical body falls through a viscous fluid, it experiences a viscous force. The magnitude of the viscous force increases with the increases in velocity of the body falling under the action of its weight. As a result, the viscous force soon balance the driving force (weight of the body) and the body starts moving with constant velocity, known as its terminal velocity.
According to Stokes, when a spherical body falls through a viscous fluid, it experiences a viscous force. The magnitude of the viscous force increases with the increases in velocity of the body falling under the action of its weight. As a result, the viscous force soon balance the driving force (weight of the body) and the body starts moving with constant velocity, known as its terminal velocity.
