The neck and bottom of a bottle are 3 cm and 15 cm in radius respectively. If the cork is pressed with a force 12N in the neck of the bottle, then force exerted on the bottom of the bottle is
A. 30N
B. 150N
C. 300N
D. 600N
Answer
629.7k+ views
Hint: Pressure is defined as the force per unit area. Different pressure is being applied on the neck and the bottom of the bottle. Using Pascal’s law, we can find the force exerted on the bottom of the bottle.
Formula used:
Pressure applied is given as:
$P = \dfrac{F}{A}$
where F is the force applied and A is the surface area.
The area of a circle is given as:
$A = \pi {r^2}$
where r is the radius of the circle.
Complete step-by-step answer:
We are given the radius of the neck and the bottom to be 3 cm and 15 cm. Therefore we can write,
$
{r_1} = 3cm = 0.03m \\
{r_2} = 15cm = 0.15m \\
$
From the radius, we can calculate the surface areas easily. We are given the force applied on the neck through the cork. Therefore, we can write
${F_1} = 12N$
We need to find the force on the bottom. First we calculate the pressure on the neck and bottom as follows:
$
{P_1} = \dfrac{{{F_1}}}{{{A_1}}} = \dfrac{{{F_1}}}{{\pi r_1^2}} \\
{P_2} = \dfrac{{{F_2}}}{{{A_2}}} = \dfrac{{{F_2}}}{{\pi r_2^2}} \\
$
Substituting the values of forces and areas known for the neck and bottom of the bottle, we get
$
{P_1} = \dfrac{{12}}{{\pi {{\left( {0.03} \right)}^2}}} \\
{P_2} = \dfrac{{{F_2}}}{{\pi {{\left( {0.15} \right)}^2}}} \\
$
Pascal’s law states that a liquid exerts pressure equally in all directions. So, the pressure exerted at the neck of the bottle is equal to pressure exerted at the bottom of the bottle. Hence, we can equate the two pressure and obtain the force on bottom of bottle as follows:
$
{P_1} = {P_2} \\
\dfrac{{12}}{{\pi {{\left( {0.03} \right)}^2}}} = \dfrac{{{F_2}}}{{\pi {{\left( {0.15} \right)}^2}}} \\
{F_2} = 12 \times {\left( {\dfrac{{0.15}}{{0.03}}} \right)^2} = 300N \\
$
Hence, the correct answer is option C.
Note: In this question, we have ignored the pressure due to height. Otherwise, there should be additional pressure at the bottom of the bottle due to the weight of the liquid. Also Pascal's law is applicable here because the liquid is static.
Formula used:
Pressure applied is given as:
$P = \dfrac{F}{A}$
where F is the force applied and A is the surface area.
The area of a circle is given as:
$A = \pi {r^2}$
where r is the radius of the circle.
Complete step-by-step answer:
We are given the radius of the neck and the bottom to be 3 cm and 15 cm. Therefore we can write,
$
{r_1} = 3cm = 0.03m \\
{r_2} = 15cm = 0.15m \\
$
From the radius, we can calculate the surface areas easily. We are given the force applied on the neck through the cork. Therefore, we can write
${F_1} = 12N$
We need to find the force on the bottom. First we calculate the pressure on the neck and bottom as follows:
$
{P_1} = \dfrac{{{F_1}}}{{{A_1}}} = \dfrac{{{F_1}}}{{\pi r_1^2}} \\
{P_2} = \dfrac{{{F_2}}}{{{A_2}}} = \dfrac{{{F_2}}}{{\pi r_2^2}} \\
$
Substituting the values of forces and areas known for the neck and bottom of the bottle, we get
$
{P_1} = \dfrac{{12}}{{\pi {{\left( {0.03} \right)}^2}}} \\
{P_2} = \dfrac{{{F_2}}}{{\pi {{\left( {0.15} \right)}^2}}} \\
$
Pascal’s law states that a liquid exerts pressure equally in all directions. So, the pressure exerted at the neck of the bottle is equal to pressure exerted at the bottom of the bottle. Hence, we can equate the two pressure and obtain the force on bottom of bottle as follows:
$
{P_1} = {P_2} \\
\dfrac{{12}}{{\pi {{\left( {0.03} \right)}^2}}} = \dfrac{{{F_2}}}{{\pi {{\left( {0.15} \right)}^2}}} \\
{F_2} = 12 \times {\left( {\dfrac{{0.15}}{{0.03}}} \right)^2} = 300N \\
$
Hence, the correct answer is option C.
Note: In this question, we have ignored the pressure due to height. Otherwise, there should be additional pressure at the bottom of the bottle due to the weight of the liquid. Also Pascal's law is applicable here because the liquid is static.
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