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. At which of the following temperatures, the value of surface tension of water is minimum?
Solution
Surface tension of water decreases with rise in temperature
Surface tension of water decreases with rise in temperature
Q2.A vessel, whose bottom has round holes with diameter 0.1 mm is filled with water. The maximum height upto which water can be filled without leakage is (Surface tension = 75 dyne cm(-1) and g=1000 cms(-2)
Solution
hρg=2S/r or h=2S/rρg =(2×75)/(0.005×1×1000)=30 cm
hρg=2S/r or h=2S/rρg =(2×75)/(0.005×1×1000)=30 cm
Q3. A horizontal pipe of cross-sectional diameter 5 cm carries water at velocity of 4 ms(-1). The pipe is connected to a smaller pipe with a cross-sectional diameter 4 cm. the velocity of water through the smaller pipe is
Solution
From the principle of continuity, Av=constant or A1 v1=A2 v2
From the principle of continuity, Av=constant or A1 v1=A2 v2
Q4. We have two (narrow) capillary tubes T1 and T2. Their lengths are l1 and l2 and radii of cross-section are r1 and r2 respectively. The rate of flow of water under a pressure difference P through tube T1 is 8cm3/sec. If l1=2l2 and r1=r2, what will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before (=P)
Solution
V=(πPr^4 )/8ηl=(8cm^3)/sec For composite tube V_1=(Pπr4)/8η(l+l/2) =2/3 (πPr4)/8ηl=2/3×8=16/3 (cm3)/sec [∵l1=l=2l2 or l2=l/2 ]
V=(πPr^4 )/8ηl=(8cm^3)/sec For composite tube V_1=(Pπr4)/8η(l+l/2) =2/3 (πPr4)/8ηl=2/3×8=16/3 (cm3)/sec [∵l1=l=2l2 or l2=l/2 ]
Q5.A glass tube of uniform internal radius r has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End 1 has a hemispherical soap bubble of radius r. End 2 has sub-hemispherical soap bubble as shown in figure. Just after opening the valve.
Solution
∆p_1=4T/r_1 and ∆p_2=4T/r_2 r1 > r2 ∴ ∆p1>∆p2 Air will flow from 1 to 2 and volume of bubble at end-1 will decreases.
∆p_1=4T/r_1 and ∆p_2=4T/r_2 r1 > r2 ∴ ∆p1>∆p2 Air will flow from 1 to 2 and volume of bubble at end-1 will decreases.
Q6. Water from a tap emerges vertically downwards with an initial speed of 1.0 ms(-1). The cross-sectional area of the tap is 〖10〗(-4) m2. Assume that the pressure is constant throughout the stream of water and that the flow is steady. The cross-sectional area of the stream 0.15 m below the tap is
Solution
Velocity of water 0.15 m below the tap is given by v22=v12+2gh =〖 (1.0)〗2+2×10×0.15 = 1+3=4 ⟹ v2=2ms(-1) Now using equation of continuity, we have a1 v1=a2 v2 a(2=) (a1 v1)/v2 = (〖10〗(-4)×1)/2=5×〖10〗(-5) m2
Velocity of water 0.15 m below the tap is given by v22=v12+2gh =〖 (1.0)〗2+2×10×0.15 = 1+3=4 ⟹ v2=2ms(-1) Now using equation of continuity, we have a1 v1=a2 v2 a(2=) (a1 v1)/v2 = (〖10〗(-4)×1)/2=5×〖10〗(-5) m2
Q7. Two substances of densities ρ_1 and ρ_2 are mixed in equal volume and the relative density of mixture is 4. When they are mixed in equal masses, the relative density of the mixture is 3. The values of ρ_1 and ρ_2 are
Solution
When substances are mixed in equal volume then density =(ρ1+ρ2)/2=4⇒ρ1+ρ2=8 …(i) When substances are mixed in equal masses the density =(2ρ1 ρ2)/(ρ1+ρ2 )=3 ⇒2ρ1 ρ2=3(ρ1+ρ2) …(ii) By solving (i) and (ii) we get ρ1=6 and ρ2=2
When substances are mixed in equal volume then density =(ρ1+ρ2)/2=4⇒ρ1+ρ2=8 …(i) When substances are mixed in equal masses the density =(2ρ1 ρ2)/(ρ1+ρ2 )=3 ⇒2ρ1 ρ2=3(ρ1+ρ2) …(ii) By solving (i) and (ii) we get ρ1=6 and ρ2=2
Q8.A tank is filled with water of density 1 g cm(-3) and oil of density 0.9 g cm(-3) . The height of water layer is 100 cm and of oil layer is 400 cm. If g=980 〖cm s〗(-2) , then the velocity of efflux from an opening in the bottom of the tank is
Solution
Let〖 d〗_w and d_0 be the densities of water and oil, then the pressure at the bottom of the tank =hw dw g+h0 d0 g Let this pressure be equivalent to pressure due to water of height h then hdw g=hw dw g+h0 d0 g ∴ h=hw+(h0 d0)/dw =100+(400×0.9)/1 =100+360=460 According to Toricelli’s theorem, v=√2gh=√(2×980×460) =√(920×980 ) cm s(-1)
Let〖 d〗_w and d_0 be the densities of water and oil, then the pressure at the bottom of the tank =hw dw g+h0 d0 g Let this pressure be equivalent to pressure due to water of height h then hdw g=hw dw g+h0 d0 g ∴ h=hw+(h0 d0)/dw =100+(400×0.9)/1 =100+360=460 According to Toricelli’s theorem, v=√2gh=√(2×980×460) =√(920×980 ) cm s(-1)
Q9.A rectangular plate 2m×3m is immersed in water in such a way that its greatest and least depth are 6m and 4m respectively from the water surface. The total thrust on the plate is
Solution
Given size of the plate =2m×5m and Greatest and least depths of the plate are 6m and 4m We know that area of the plate A=2×3=6m2 And depth of centre of the plate x-=(6+4)/2=5 m ∴ Total thrust on the plate ρ=ρw g A ⃗x =〖10〗3×9.8×6×5 =294×〖10〗3 N
Given size of the plate =2m×5m and Greatest and least depths of the plate are 6m and 4m We know that area of the plate A=2×3=6m2 And depth of centre of the plate x-=(6+4)/2=5 m ∴ Total thrust on the plate ρ=ρw g A ⃗x =〖10〗3×9.8×6×5 =294×〖10〗3 N
Q10. A cylinder is filled with liquid of density d upto a height h. If the cylinder is at rest, then the mean pressure of the walls is
Solution
Pressure at the bottom =hdg Pressure at the top due to liquid column =0 ∴ Mean pressure=(hdg+0)/2=hdg/2
Pressure at the bottom =hdg Pressure at the top due to liquid column =0 ∴ Mean pressure=(hdg+0)/2=hdg/2
