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.
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Q1. A uniform tapering vessel shown in figure is filled with liquid of density 900 kgm(-3). The force that acts on the base of the vessel due to liquid is (take g =10 ms(-2))
Solution
Force on the base of the vessel = pressure ×area of the base =h ρ g×A=0.4×900×10×2× 10(-3) =7.2 N
Force on the base of the vessel = pressure ×area of the base =h ρ g×A=0.4×900×10×2× 10(-3) =7.2 N
Q2.Two capillary tubes of radii 0.2 cm and 0.4 cm are dipped in the same liquid. The ratio of heights through which liquid will rise in the tubes is
Solution
Height, h∝1/R So h1/h2=R2/R1=0.4/0.2=2
Height, h∝1/R So h1/h2=R2/R1=0.4/0.2=2
Q3. The water flows from a tap of diameter 1.25 cm with a rate of 5× 10(-5) m3 s(-1). The density and coefficient of viscosity of water are 103 kg m(-3) and 10(-3) Pas, respectively. The flow of water is
Solution
Here, diameter D=1.25cm=1.25×〖10〗(-2) m Density of water ρ=〖10〗3 kgm(-3) Coefficient of viscosity η=〖10〗(-3) Pas Rate of flow of water Q=5×〖10〗(-5) m3 s(-1) Reynold’s number N_R=vρD/η …(i) Where v is the speed of flow Rate of flow of water Q= Area of cross section × speed of flow Q=(πD2)/4×v⇒v=4Q/(πD2 ) Substituting the value of v in eqn. (i), we get N_R=4QρD/(πD2 η)=4Qρ/πDη Substituting the values, we get N_R=(4×5×〖10〗(-5)×〖10〗3)/((22/7)×1.25×〖10〗(-2)×〖10〗(-3) )≈5100 For N_R>3000, the flow is turbulent Hence, the flow of water is turbulent with Reynold’s number 5100
Here, diameter D=1.25cm=1.25×〖10〗(-2) m Density of water ρ=〖10〗3 kgm(-3) Coefficient of viscosity η=〖10〗(-3) Pas Rate of flow of water Q=5×〖10〗(-5) m3 s(-1) Reynold’s number N_R=vρD/η …(i) Where v is the speed of flow Rate of flow of water Q= Area of cross section × speed of flow Q=(πD2)/4×v⇒v=4Q/(πD2 ) Substituting the value of v in eqn. (i), we get N_R=4QρD/(πD2 η)=4Qρ/πDη Substituting the values, we get N_R=(4×5×〖10〗(-5)×〖10〗3)/((22/7)×1.25×〖10〗(-2)×〖10〗(-3) )≈5100 For N_R>3000, the flow is turbulent Hence, the flow of water is turbulent with Reynold’s number 5100
Q4. If there were no gravity, which of the following will not be there for fluid?
Solution
Archimedes’ upward thrust will be absent for a fluid, if there were no gravity.
Archimedes’ upward thrust will be absent for a fluid, if there were no gravity.
Q5. A cylindrical tank has a hole of 1 cm^2 in its bottom. If the water is allowed to flow into the tank from a tube above it at the rate of 70 cm3/sec. then the maximum height up to which water can rise in the tank is
Solution
The height of water in the tank becomes maximum when the volume of water flowing into the tank per second becomes equal to the volume flowing out per second. Volume of water flowing out per second =Av=A√2gh …(i) Volume of water flowing in per second =70 cm3/sec …(ii) From (i) and (ii) we get A√2gh=70⇒1×√2gh=70⇒1×√(2×980×h)=70 ∴h=4900/1960=2.5 cm
The height of water in the tank becomes maximum when the volume of water flowing into the tank per second becomes equal to the volume flowing out per second. Volume of water flowing out per second =Av=A√2gh …(i) Volume of water flowing in per second =70 cm3/sec …(ii) From (i) and (ii) we get A√2gh=70⇒1×√2gh=70⇒1×√(2×980×h)=70 ∴h=4900/1960=2.5 cm
Q6. The spring balance A reads 2 kg with a block of mass m suspended from it. A balance B reads 5 kg when a beaker with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid in a beaker as shown in figure
Solution
The effective weight of the block in liquid will become less than 2 kg due to buoyancy of liquid. As a result of which A will read less than 2 kg As the body immersed in liquid has some effective weight acting downwards so the reading of B will be more than 5 kg
The effective weight of the block in liquid will become less than 2 kg due to buoyancy of liquid. As a result of which A will read less than 2 kg As the body immersed in liquid has some effective weight acting downwards so the reading of B will be more than 5 kg
Q7.If the rise in height of capillary of two tubes are 6.6 cm and 2.2 cm, then the ratio of the radii of tubes is
Solution
1:3
1:3
Q8.A closed rectangular tank is completely filled with water and is accelerated horizontally with an acceleration a towards right. Pressure is (i) maximum at, and (ii) minimum at
Solution
Q9. Two pieces of glass plate one upon the other with a little water in between them cannot be separated easily because of
Solution
The force of surface tension pulls the plates towards each other
The force of surface tension pulls the plates towards each other
Q10. Two rain drops falling through air have radii in the ratio 1:2. They will have terminal velocity in the ratio
Solution
1:4
1:4
