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.**A U-tube is partially filled with water. Oil which does not mix with water is next poured into one side until water rises by 25 cm. On the other side, if the density of oil be 0.8, the oil level will stand higher than the water level by

Solution

As the water level rises in one arm; it falls in another arm by 25 m. Equating the pressure at depth 50 cm down in arm of water with other due to liquid, we have h×0.8×g=50×1×g Or h=50/0.8=62.5 cm Height of oil in one limb above the water in another limb =62.5-50=12.5 cm

As the water level rises in one arm; it falls in another arm by 25 m. Equating the pressure at depth 50 cm down in arm of water with other due to liquid, we have h×0.8×g=50×1×g Or h=50/0.8=62.5 cm Height of oil in one limb above the water in another limb =62.5-50=12.5 cm

**Q2.**A tank of height H is fully filled with water. If the water rushing from a hole made in the tank below the free surface, strikes the floor at maximum horizontal distance, then the depth of the hole from the free surface must be

Solution

As hole is made in the tank below the free surface of water, so water rushing from this hole follows a parabolic path. The velocity of efflux liquid, v=√2gh Time t=√(2(H-h)/g) Horizontal range, R=vt R=(2gh×2(H-h)/g)

As hole is made in the tank below the free surface of water, so water rushing from this hole follows a parabolic path. The velocity of efflux liquid, v=√2gh Time t=√(2(H-h)/g) Horizontal range, R=vt R=(2gh×2(H-h)/g)

^{(1/2)}ie,R^{2}=4h(H-h)=4(Hh-h^{2}) The range is maximum if dR/dh=0 or 2RdR/dh=4(H-2h) or 0=(H-2h) or h=H/2**Q3.**In a turbulent flow, the velocity of the liquid in contact with the walls of the tube is

Solution

In a turbulent flow, the velocity of the liquid in contact with the walls of the tube is equal is critical velocity.

In a turbulent flow, the velocity of the liquid in contact with the walls of the tube is equal is critical velocity.

**Q4.**Two mercury drop (each of radius r) merge to form a bigger drop, if T is the surface tension is

Solution

Let R be the radius of the bigger drop, then Volume of bigger drop=2×volume of small drop 4/3 πR

Let R be the radius of the bigger drop, then Volume of bigger drop=2×volume of small drop 4/3 πR

^{3}=2×4/3 πr^{3}R=2^{(1/3)}r Surface energy of bigger drop, E=4πR^{2}T =4×2^{(2/3)}πr^{2}T =2^{2}πr^{2}T**Q5.**A ball of mass m and radius r is released in a viscous liquid. The value of its terminal velocity is proportional

Solution

v=(2r

v=(2r

^{2}ρg)/9η r⇒v∝r^{2}ρ But mass, m=4/3 πr^{3}ρ or ρ∝m/r^{3}; Hence, v∝r^{2}(m/r^{3}) or v∝m/r**Q6.**16 cm^3 Of water flows per sec through a capillary tube of radius a cm and of length l cm when connected to a pressure head of h cm of water. If a tube of the same length and radius a/2 cm is connected to the same pressure head, the quantity of water flowing through the tube per second will be

Solution

V=(π p r

V=(π p r

^{4})/(8 η l),ie,V∝r^{4}V^{'}/V=(a/2)^{4}/a^{4}=1/16 or V^{'}=V/16=16/16=1 cm^{3}**Q7.**A drop of water breaks into two droplets of equal size. In this process, which of the following statements is correct?

Solution

When two drops are splitted, the law of conservation of mass is obeyed

When two drops are splitted, the law of conservation of mass is obeyed

**Q8.**The excess pressure inside a spherical drop of radius r of a liquid of surface tension T is

Solution

The excess pressure inside a liquid drop is ∆p=2T/r or ∆p∝T/r

The excess pressure inside a liquid drop is ∆p=2T/r or ∆p∝T/r

**Q9.**A trough contains mercury to a depth of 3.6 cm. If some amount of mercury is poured in it then height of mercury in the trough will be

Solution

Let A be the area of cross-section of through and ρ be the density of mercury Initial mass of mercury in trough =A×3.6×ρ Final mass of mercury in trough =A h

Let A be the area of cross-section of through and ρ be the density of mercury Initial mass of mercury in trough =A×3.6×ρ Final mass of mercury in trough =A h

^{'}ρ=(A×3.6×ρ)×2 or h^{'}=7.2 cm**Q10.**Surface tension of a liquid is due to

Solution

Surface tension of a liquid is due to force of attraction between like molecules of a liquid ie cohesive force between the molecules

Surface tension of a liquid is due to force of attraction between like molecules of a liquid ie cohesive force between the molecules