NEET Test Series from KOTA - 10 Papers In MS WORD
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Mechanical Properties of Fluids
143167
A hydraulic lift can lift a maximum load of $3000 \mathrm{~kg}$ wt. The area of cross section of the piston carrying the load is $4.25 \times 10^{-2} \mathrm{~m}^{2}$. The maximum pressure the smaller piston would have to bear is
A Given, Hydraulic lift can lift a maximum load $=3000 \mathrm{~kg}$ Cross - sectional area of a large piston $=4.25 \times 10^{-2} \mathrm{~m}^{2}$ Now, Applying Pascal's law, The maximum pressure exerted on the load-carrying piston, $\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}$ $=\frac{3000 \times \mathrm{g}}{4.25 \times 10^{-2}}$ Pressure, $\mathrm{P}=\frac{3000 \times 9.81}{4.25 \times 10^{-2}}$ $=6.92 \times 10^{5} \mathrm{~N} \cdot \mathrm{m}^{-2}$
AP EAMCET-23.09.2020
Mechanical Properties of Fluids
143169
The acceleration ' $a$ ' of the vertical U-tube is
1 $\mathrm{g}$
2 $\frac{\mathrm{g}}{2}$
3 $2 \mathrm{~g}$
4 0
Explanation:
B From figure- $\tan \theta=\frac{\mathrm{a}}{\mathrm{g}}$ Now, as shown in vertical U-tube $\tan \theta=\frac{2 b}{4 b}$ $\tan \theta=1 / 2$ From the equation (i) and (ii) $\frac{\mathrm{a}}{\mathrm{g}}=\frac{1}{2}$ $\mathrm{a}=\frac{\mathrm{g}}{2}$
AP EAMCET-24.09.2020
Mechanical Properties of Fluids
143170
The pressure on a swimmer $20 \mathrm{~m}$ below the surface of water at sea level is
1 $1.0 \mathrm{~atm}$
2 $2.0 \mathrm{~atm}$
3 $2.5 \mathrm{~atm}$
4 $3.0 \mathrm{~atm}$
Explanation:
D Given, Height of swimmer below the surface of water $(\mathrm{h})=$ $20 \mathrm{~m}$ We know, $\text { Density of water }=1000 \mathrm{kgm}^{-3}$ and Pressure $(\mathrm{P})=\rho \mathrm{gh}$ Then, $\quad P=1000 \times 10 \times 20=2 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$ $\mathrm{P}=2 \mathrm{~atm}$ Now, Pressure on a swimmer $=\mathrm{P}_{\mathrm{atm}}+\mathrm{P}$ $=1 \mathrm{~atm}+2 \mathrm{~atm}$ $=3 \mathrm{~atm}$
AMU-2010
Mechanical Properties of Fluids
143171
The gauge pressure produced by a machine to suck mud of density $1800 \mathrm{~kg} / \mathrm{m}^{3}$ up a tube by a height of $2.0 \mathrm{~m}$ from ground is (Take $\mathrm{g}=\mathbf{9 . 8}$ $\mathbf{m} / \mathbf{s}^{2}$ )
1 $2.0 \times 10^{4} \mathrm{~Pa}$
2 $3.5 \times 10^{4} \mathrm{~Pa}$
3 $6.2 \times 10^{4} \mathrm{~Pa}$
4 $9.8 \times 10^{4} \mathrm{~Pa}$
Explanation:
B Given, $\text { Density }(\rho)=1800 \mathrm{~kg} / \mathrm{m}^{3}$ $\text { Height }(\mathrm{h})=-2 \mathrm{~m}$ Acceleration due to gravity $(\mathrm{g})=9.8 \mathrm{~m} / \mathrm{s}^{2}$ We know, Gauge Pressure $(\mathrm{P})=\rho \mathrm{gh}$ $\mathrm{P}=1800 \times(2) \times 9.8$ $\mathrm{P}=3528.0 \mathrm{~Pa}$ $\mathrm{P}=3.5 \times 10^{4} \mathrm{~Pa}$
143167
A hydraulic lift can lift a maximum load of $3000 \mathrm{~kg}$ wt. The area of cross section of the piston carrying the load is $4.25 \times 10^{-2} \mathrm{~m}^{2}$. The maximum pressure the smaller piston would have to bear is
A Given, Hydraulic lift can lift a maximum load $=3000 \mathrm{~kg}$ Cross - sectional area of a large piston $=4.25 \times 10^{-2} \mathrm{~m}^{2}$ Now, Applying Pascal's law, The maximum pressure exerted on the load-carrying piston, $\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}$ $=\frac{3000 \times \mathrm{g}}{4.25 \times 10^{-2}}$ Pressure, $\mathrm{P}=\frac{3000 \times 9.81}{4.25 \times 10^{-2}}$ $=6.92 \times 10^{5} \mathrm{~N} \cdot \mathrm{m}^{-2}$
AP EAMCET-23.09.2020
Mechanical Properties of Fluids
143169
The acceleration ' $a$ ' of the vertical U-tube is
1 $\mathrm{g}$
2 $\frac{\mathrm{g}}{2}$
3 $2 \mathrm{~g}$
4 0
Explanation:
B From figure- $\tan \theta=\frac{\mathrm{a}}{\mathrm{g}}$ Now, as shown in vertical U-tube $\tan \theta=\frac{2 b}{4 b}$ $\tan \theta=1 / 2$ From the equation (i) and (ii) $\frac{\mathrm{a}}{\mathrm{g}}=\frac{1}{2}$ $\mathrm{a}=\frac{\mathrm{g}}{2}$
AP EAMCET-24.09.2020
Mechanical Properties of Fluids
143170
The pressure on a swimmer $20 \mathrm{~m}$ below the surface of water at sea level is
1 $1.0 \mathrm{~atm}$
2 $2.0 \mathrm{~atm}$
3 $2.5 \mathrm{~atm}$
4 $3.0 \mathrm{~atm}$
Explanation:
D Given, Height of swimmer below the surface of water $(\mathrm{h})=$ $20 \mathrm{~m}$ We know, $\text { Density of water }=1000 \mathrm{kgm}^{-3}$ and Pressure $(\mathrm{P})=\rho \mathrm{gh}$ Then, $\quad P=1000 \times 10 \times 20=2 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$ $\mathrm{P}=2 \mathrm{~atm}$ Now, Pressure on a swimmer $=\mathrm{P}_{\mathrm{atm}}+\mathrm{P}$ $=1 \mathrm{~atm}+2 \mathrm{~atm}$ $=3 \mathrm{~atm}$
AMU-2010
Mechanical Properties of Fluids
143171
The gauge pressure produced by a machine to suck mud of density $1800 \mathrm{~kg} / \mathrm{m}^{3}$ up a tube by a height of $2.0 \mathrm{~m}$ from ground is (Take $\mathrm{g}=\mathbf{9 . 8}$ $\mathbf{m} / \mathbf{s}^{2}$ )
1 $2.0 \times 10^{4} \mathrm{~Pa}$
2 $3.5 \times 10^{4} \mathrm{~Pa}$
3 $6.2 \times 10^{4} \mathrm{~Pa}$
4 $9.8 \times 10^{4} \mathrm{~Pa}$
Explanation:
B Given, $\text { Density }(\rho)=1800 \mathrm{~kg} / \mathrm{m}^{3}$ $\text { Height }(\mathrm{h})=-2 \mathrm{~m}$ Acceleration due to gravity $(\mathrm{g})=9.8 \mathrm{~m} / \mathrm{s}^{2}$ We know, Gauge Pressure $(\mathrm{P})=\rho \mathrm{gh}$ $\mathrm{P}=1800 \times(2) \times 9.8$ $\mathrm{P}=3528.0 \mathrm{~Pa}$ $\mathrm{P}=3.5 \times 10^{4} \mathrm{~Pa}$
143167
A hydraulic lift can lift a maximum load of $3000 \mathrm{~kg}$ wt. The area of cross section of the piston carrying the load is $4.25 \times 10^{-2} \mathrm{~m}^{2}$. The maximum pressure the smaller piston would have to bear is
A Given, Hydraulic lift can lift a maximum load $=3000 \mathrm{~kg}$ Cross - sectional area of a large piston $=4.25 \times 10^{-2} \mathrm{~m}^{2}$ Now, Applying Pascal's law, The maximum pressure exerted on the load-carrying piston, $\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}$ $=\frac{3000 \times \mathrm{g}}{4.25 \times 10^{-2}}$ Pressure, $\mathrm{P}=\frac{3000 \times 9.81}{4.25 \times 10^{-2}}$ $=6.92 \times 10^{5} \mathrm{~N} \cdot \mathrm{m}^{-2}$
AP EAMCET-23.09.2020
Mechanical Properties of Fluids
143169
The acceleration ' $a$ ' of the vertical U-tube is
1 $\mathrm{g}$
2 $\frac{\mathrm{g}}{2}$
3 $2 \mathrm{~g}$
4 0
Explanation:
B From figure- $\tan \theta=\frac{\mathrm{a}}{\mathrm{g}}$ Now, as shown in vertical U-tube $\tan \theta=\frac{2 b}{4 b}$ $\tan \theta=1 / 2$ From the equation (i) and (ii) $\frac{\mathrm{a}}{\mathrm{g}}=\frac{1}{2}$ $\mathrm{a}=\frac{\mathrm{g}}{2}$
AP EAMCET-24.09.2020
Mechanical Properties of Fluids
143170
The pressure on a swimmer $20 \mathrm{~m}$ below the surface of water at sea level is
1 $1.0 \mathrm{~atm}$
2 $2.0 \mathrm{~atm}$
3 $2.5 \mathrm{~atm}$
4 $3.0 \mathrm{~atm}$
Explanation:
D Given, Height of swimmer below the surface of water $(\mathrm{h})=$ $20 \mathrm{~m}$ We know, $\text { Density of water }=1000 \mathrm{kgm}^{-3}$ and Pressure $(\mathrm{P})=\rho \mathrm{gh}$ Then, $\quad P=1000 \times 10 \times 20=2 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$ $\mathrm{P}=2 \mathrm{~atm}$ Now, Pressure on a swimmer $=\mathrm{P}_{\mathrm{atm}}+\mathrm{P}$ $=1 \mathrm{~atm}+2 \mathrm{~atm}$ $=3 \mathrm{~atm}$
AMU-2010
Mechanical Properties of Fluids
143171
The gauge pressure produced by a machine to suck mud of density $1800 \mathrm{~kg} / \mathrm{m}^{3}$ up a tube by a height of $2.0 \mathrm{~m}$ from ground is (Take $\mathrm{g}=\mathbf{9 . 8}$ $\mathbf{m} / \mathbf{s}^{2}$ )
1 $2.0 \times 10^{4} \mathrm{~Pa}$
2 $3.5 \times 10^{4} \mathrm{~Pa}$
3 $6.2 \times 10^{4} \mathrm{~Pa}$
4 $9.8 \times 10^{4} \mathrm{~Pa}$
Explanation:
B Given, $\text { Density }(\rho)=1800 \mathrm{~kg} / \mathrm{m}^{3}$ $\text { Height }(\mathrm{h})=-2 \mathrm{~m}$ Acceleration due to gravity $(\mathrm{g})=9.8 \mathrm{~m} / \mathrm{s}^{2}$ We know, Gauge Pressure $(\mathrm{P})=\rho \mathrm{gh}$ $\mathrm{P}=1800 \times(2) \times 9.8$ $\mathrm{P}=3528.0 \mathrm{~Pa}$ $\mathrm{P}=3.5 \times 10^{4} \mathrm{~Pa}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Mechanical Properties of Fluids
143167
A hydraulic lift can lift a maximum load of $3000 \mathrm{~kg}$ wt. The area of cross section of the piston carrying the load is $4.25 \times 10^{-2} \mathrm{~m}^{2}$. The maximum pressure the smaller piston would have to bear is
A Given, Hydraulic lift can lift a maximum load $=3000 \mathrm{~kg}$ Cross - sectional area of a large piston $=4.25 \times 10^{-2} \mathrm{~m}^{2}$ Now, Applying Pascal's law, The maximum pressure exerted on the load-carrying piston, $\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}$ $=\frac{3000 \times \mathrm{g}}{4.25 \times 10^{-2}}$ Pressure, $\mathrm{P}=\frac{3000 \times 9.81}{4.25 \times 10^{-2}}$ $=6.92 \times 10^{5} \mathrm{~N} \cdot \mathrm{m}^{-2}$
AP EAMCET-23.09.2020
Mechanical Properties of Fluids
143169
The acceleration ' $a$ ' of the vertical U-tube is
1 $\mathrm{g}$
2 $\frac{\mathrm{g}}{2}$
3 $2 \mathrm{~g}$
4 0
Explanation:
B From figure- $\tan \theta=\frac{\mathrm{a}}{\mathrm{g}}$ Now, as shown in vertical U-tube $\tan \theta=\frac{2 b}{4 b}$ $\tan \theta=1 / 2$ From the equation (i) and (ii) $\frac{\mathrm{a}}{\mathrm{g}}=\frac{1}{2}$ $\mathrm{a}=\frac{\mathrm{g}}{2}$
AP EAMCET-24.09.2020
Mechanical Properties of Fluids
143170
The pressure on a swimmer $20 \mathrm{~m}$ below the surface of water at sea level is
1 $1.0 \mathrm{~atm}$
2 $2.0 \mathrm{~atm}$
3 $2.5 \mathrm{~atm}$
4 $3.0 \mathrm{~atm}$
Explanation:
D Given, Height of swimmer below the surface of water $(\mathrm{h})=$ $20 \mathrm{~m}$ We know, $\text { Density of water }=1000 \mathrm{kgm}^{-3}$ and Pressure $(\mathrm{P})=\rho \mathrm{gh}$ Then, $\quad P=1000 \times 10 \times 20=2 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$ $\mathrm{P}=2 \mathrm{~atm}$ Now, Pressure on a swimmer $=\mathrm{P}_{\mathrm{atm}}+\mathrm{P}$ $=1 \mathrm{~atm}+2 \mathrm{~atm}$ $=3 \mathrm{~atm}$
AMU-2010
Mechanical Properties of Fluids
143171
The gauge pressure produced by a machine to suck mud of density $1800 \mathrm{~kg} / \mathrm{m}^{3}$ up a tube by a height of $2.0 \mathrm{~m}$ from ground is (Take $\mathrm{g}=\mathbf{9 . 8}$ $\mathbf{m} / \mathbf{s}^{2}$ )
1 $2.0 \times 10^{4} \mathrm{~Pa}$
2 $3.5 \times 10^{4} \mathrm{~Pa}$
3 $6.2 \times 10^{4} \mathrm{~Pa}$
4 $9.8 \times 10^{4} \mathrm{~Pa}$
Explanation:
B Given, $\text { Density }(\rho)=1800 \mathrm{~kg} / \mathrm{m}^{3}$ $\text { Height }(\mathrm{h})=-2 \mathrm{~m}$ Acceleration due to gravity $(\mathrm{g})=9.8 \mathrm{~m} / \mathrm{s}^{2}$ We know, Gauge Pressure $(\mathrm{P})=\rho \mathrm{gh}$ $\mathrm{P}=1800 \times(2) \times 9.8$ $\mathrm{P}=3528.0 \mathrm{~Pa}$ $\mathrm{P}=3.5 \times 10^{4} \mathrm{~Pa}$
