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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. If the work done in blowing a bubble of volume V is W, then the work done in blowing a soap bubble of volume 2V will be
•  W
•  2W
•  √2W
•  41/3W
Solution
We know that surface tension is related to work as W=T×∆A Since, surface area of sphere is 4πR2and there are two free surfaces, we have W=T×8R2 ……(i) and volume of sphere=4/3 πR3 ie,V=4/3 πR3 ⟹ R=(3V/4π)(1/3) …..(ii) From Eqs. (i) and (ii), we get W=T×8π×(3V/4π)(2/3) ⟹ W∝V(2/3) ∴ W1∝V_2(2/3) and W2∝V2(2/3) W22/W1 =(〖2V〗1/V1 )(2/3) ⟹ W2=2(2/3) W1=4(1/3)W

Q2.A cylinder drum, open at the top, contains 15 L of water. It drains out through a small opening at the bottom. 5 L of water comes out in time t1, the next 5 L in further time t2 and the last 5 L in further time t3. Then :
•  t1<t2<t3
•  t1>t2>t3
•  t1=t2=t3
•  t1>t2=t3
Solution
If h is the initial height of liquid in drum above the small opening, then velocity of efflux, v=√2gh. As the water drains out, h decreases, hence v decreases. This reduces the rate of drainage of water. Due to which, as the drainage continues, a longer time is required to drain out the same volume of water

Q3.  A weightless bag is filled with 5 kg of water and then weighed in water. The reading of spring balance is
•  5kg
•  2.5kg
•  1.25kg
•  zero
Solution
Since, weight of bag with water is equal to the weight of water displaced, hence reading of spring balance is zero

Q4. With an increase in temperature, surface tension of liquid (except molten copper and cadmium)
•  increases
•  remain same
•  decreases
•  first decreases and then increases
Solution
The surface tension of liquid decreases with rise of temperature. The surface tension of liquid is zero at its boiling point and it vanishes at critical temperature. At critical temperature intermolecular forces for liquid and gases becomes equal and liquid can expand without any restriction. For small temperature differences, the variation in surface tension with temperature is linear and is given by relation T1=T0 (1-αt) Where T1,T0 are the surface tension at t℃ and 0℃ respectively and α is the temperature coefficient of surface tension.

Q5.&nbspIf the length of tube is less and cannot accommodate the maximum rise of liquid then
•  liquid will form fountain
•  liquid will not rise
•  the meniscus will adjust itself so that the water does not spill
•  none of the above
Solution
If the length of the tube h' is less than h, is found that the liquid dose not overflow. In a tube of insufficient length, the liquid rises upto the top of the tube and increases the radius of curvature, of its mentiscus to a value R ,so that R' h'=Rh
ie, smaller the length (h') of the tube, greater will be the radius of curvature (R') of the meniscus, but the liquid will never overflow.

Q6.  Two capillaries of radii r1 andr2, length l1 and l2 respectively are in series. A liquid of viscosity η is flowing through the combination under a pressure difference p. What is the rate of volume flow of liquid?
•  (π p)/(8 η) (l4/r14)+(l4/r24)-1
•  (8π p)/η (l1/r14 )+(l2/r24)
• (π p)/(8 η) (r14/l1) +(r24/l2 )-1
•  (π p)/(8 η) (l1/r14 )+l2/r24)-1
Solution
The rate of flow of liquid (V) through capillary tube is V=(π p r4)/8ηl=p((π r4)/8ηl)=p/R=(pressure difference)/resistance Where, R=8ηhl/(π r4 ) When two tubes are in series Total resistance R=R1+R2 Rate of flow of liquid, V'=p/(R1+R2 ) =p/(8η/π [l1/(r14 )+l2/(r24 )] )=πp/8η [l1/(r14 )+l2/(r24 )](-1)

Q7. Two spherical soap bubbles of radii r1 and r2 in vacuum combine under isothermal conditions. The resulting bubble has radius equal to
•  (r1+r2)/2
•  (r1 r2)/(r1+r2 )
•  √(r1 r2 )
•  √(r12+r22 )
Solution
Excess of pressure, inside the first bubble p_1=4T/r_(1 ) Similarly,p2=4T/r(2) Let the radius of the large bubble be R. then, excess of pressure inside the large bubble p=4T/R Under isothermal condition, temperature remains constant. So, pV=p1 V1+p2 V2 4T/R (4/3 πr3 )=4T/r1 (4/3 πr13)+4T/r(2) (4/3 πr23 ) R2=r12+r22 ⟹ R=√(r12+r22 )

Q8.  A marble of mass x and diameter 2 r is gently released a tall cylinder containing honey. If the marble displaces mass y(
•  (x+y)
•  T(x-y)
•  (x+y)/r
•  ((x-y))/r
Solution
If v is the terminal velocity, then xg-yg=6 π η r v Or v=((x-y))/r g/(6 π η) Or v∝((x-y))/r

Q9.  A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at the sides (figure). If the radius of the vessel is 0.05 m and the sped of rotation is 2 rad s(-1), find the difference in the height of the liquid at the centre of the vessel and its sides
•  20cm
•  4cm
•  2cm
•  0.2cm
Solution
According to Bernoulli’s Theorem; p=1/2 ρ v2 =constant. Near the ends, the velocity of liquid is higher so that pressure is lower as a result the liquid rises at the sides to compensate for this drop of pressure ie,ρ g h=1/2 ρ v2=1/2 ρr2 ω2 Hence, h=(r2 ω2)/2g=(r2 (2πv)2)/2g=(2π2 r2 v2)/g =(2×π2×(0.05)2×22)/9.8 =0.02 m=2 cm

Q10. Water flowing out of the mouth of a tap and falling vertically in streamline flow forms a tapering column, ie the area of cross-section of the liquid column decreases as it moves down. Which of the following is the most accurate explanation for this?
•  Falling water tries to reach a terminal velocity and hence, reduces the area of cross-section to balance upward and downward forces
•  As the water moves down, its speed increases and hence, its pressure decreases. It is then compressed by atmosphere
•  The surface tension causes the exposed surface area of the liquid to decrease continuously
• The mass of water flowing out per second through any cross-section must remain constant. As the water is almost incompressible, so the volume of water flowing out per second must remain constant. As this is equal to velocity × area, the area decreases as velocity increases
Solution
According to equation of continuity av= constant. As v increases, a decreases ## Want to know more

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