Question

A uniform thin disc of diameter 2R with a hole of diameter R (passing though centre of the disc) is held vertically in an incompressible liquid of density $\rho$. The top of the disc is at depth of 'R' from the free surface of the liquid. Atmospheric pressure is $P_0$. Force on the facing side of the disc (into the paper) due to liquid is $(P_0 + x\rho gr)y\pi R^2$ where x is rational and y is an integer .If $x = 13/n$ What is the value of n ?

• A. 26
• B. 13
• C. 2
• D. 1

Answer 1

Answer: (A)

26

Answer 2

The pressure at a depth $h$ is $P(h) = P_0 + \rho g h$. The disc has a hole of radius $R/2$ and outer radius $R$. The area of the disc material is $A = \pi R^2 - \pi (R/2)^2 = \frac{3}{4}\pi R^2$. The disc extends from depth $R$ to $3R$. The centroid of the disc material is at its geometric center, which is at a depth of $R + R = 2R$. The force on the facing side of the disc is given by $F = \int P dA$. We can write this as $F = P_{avg} \times A$, where $P_{avg}$ is the average pressure over the area. The average pressure is $P_{avg} = P_0 + \rho g h_{centroid} = P_0 + \rho g (2R)$. So, the force is $F = (P_0 + 2\rho g R) \times (\frac{3}{4}\pi R^2)$. $F = \frac{3}{4}P_0 \pi R^2 + \frac{3}{2}\rho g R \pi R^2$.

The problem states that the force is given by $(P_0 + x\rho gr)y\pi R^2$. Let's assume $r=R$. Then $F = (P_0 + x\rho g R) y\pi R^2 = y P_0 \pi R^2 + yx \rho g R \pi R^2$.

Comparing the two expressions for force: $y P_0 \pi R^2 + yx \rho g R \pi R^2 = \frac{3}{4}P_0 \pi R^2 + \frac{3}{2}\rho g R \pi R^2$.

Equating the coefficients of $P_0$: $y = \frac{3}{4}$. This contradicts the condition that $y$ is an integer.

Let's re-examine the given formula structure: $(P_0 + x\rho gr)y\pi R^2$. It is possible that the 'y' factor is meant to adjust the area, and the pressure term is correct. If we assume the pressure term is $P_0 + x\rho g R$ and the effective area is $A_{eff} = y\pi R^2$. Then $F = (P_0 + x\rho g R) \times A_{eff}$. From our calculation, $F = (P_0 + 2\rho g R) \times (\frac{3}{4}\pi R^2)$. Comparing these, we can infer: $x\rho g R = 2\rho g R \implies x = 2$. And $A_{eff} = y\pi R^2 = \frac{3}{4}\pi R^2 \implies y = \frac{3}{4}$.

Since $y$ must be an integer, let's assume there is a slight variation in how the formula is interpreted. If the formula is $F = (P_0 + x\rho g r) \times A$, and $A$ is the area of the disc material, $A = \frac{3}{4}\pi R^2$. Then $F = (P_0 + x\rho g r) \times \frac{3}{4}\pi R^2$. Let $r=R$. $F = (P_0 + x\rho g R) \frac{3}{4}\pi R^2 = \frac{3}{4}P_0 \pi R^2 + \frac{3}{4}x\rho g R \pi R^2$. Comparing this with our calculated force: $F = \frac{3}{4}P_0 \pi R^2 + \frac{3}{2}\rho g R \pi R^2$. Equating the coefficients of $\rho g R$: $\frac{3}{4}x = \frac{3}{2}$ $x = \frac{3}{2} \times \frac{4}{3} = 2$.

Now, let's consider the given formula $(P_0 + x\rho gr)y\pi R^2$. If we assume $x=2$ and $r=R$, the formula becomes $(P_0 + 2\rho g R)y\pi R^2$. For this to equal our calculated force $F = \frac{3}{4}P_0 \pi R^2 + \frac{3}{2}\rho g R \pi R^2$, we need: $(P_0 + 2\rho g R)y\pi R^2 = (\frac{3}{4}P_0 + \frac{3}{2}\rho g R) \pi R^2$. Dividing by $\pi R^2$: $(P_0 + 2\rho g R)y = \frac{3}{4}P_0 + \frac{3}{2}\rho g R$. $y P_0 + 2y\rho g R = \frac{3}{4}P_0 + \frac{3}{2}\rho g R$. This implies $y = 3/4$ and $2y = 3/2$, which is consistent ($y=3/4$).

However, the problem states $y$ is an integer. This implies a mismatch in the interpretation or a typo in the question. Let's assume the question intends to match the $\rho g R$ term. We have $F = \frac{3}{4}P_0 \pi R^2 + \frac{3}{2}\rho g R \pi R^2$. The given form is $F = y P_0 \pi R^2 + yx \rho g R \pi R^2$. If we assume the $\rho g R$ term is the primary match, and $y$ must be an integer. Let's try to match the $\rho g R$ term: $yx = 3/2$. And the $P_0$ term: $y = 3/4$. This still leads to $y=3/4$.

Let's consider the possibility that the 'y' is not a direct multiplier of the entire expression, but the structure implies a specific form. Let's consider the possibility that the formula is meant to be interpreted as $F = y \times (\text{Pressure}) \times (\text{Area})$. If the area is $\pi R^2$, then $F = y(P_0 + x\rho gr)\pi R^2$. $F = y P_0 \pi R^2 + yx \rho g r \pi R^2$. Let $r=R$. $F = y P_0 \pi R^2 + yx \rho g R \pi R^2$. Comparing with $F = \frac{3}{4}P_0 \pi R^2 + \frac{3}{2}\rho g R \pi R^2$. We need $y = 3/4$, which is not an integer.

Let's assume the problem meant that the force is given by $F = (P_0 + x\rho g R) \times (\text{Area})$. And the Area is given by $y\pi R^2$. So $F = (P_0 + x\rho g R) \times (y\pi R^2)$. We calculated $F = (P_0 + 2\rho g R) \times (\frac{3}{4}\pi R^2)$. For these to match, we need $x=2$ and $y=3/4$.

Given the constraint that $y$ is an integer, and $x=13/n$. Let's assume the problem implies that the coefficient of $\rho g R$ in our calculated force is $3/2$. And in the given formula, it is $yx$. So, $yx = 3/2$. And the coefficient of $P_0$ is $y$. So, $y = 3/4$. This is the contradiction.

Let's assume the problem setter intended for $y$ to be an integer that represents some factor related to the area, and the pressure term is correct. If we assume the formula is $F = (\text{Pressure}) \times (\text{Area})$. Pressure $= P_0 + x\rho gr$. Area $= y\pi R^2$. We calculated $F = (\frac{3}{4}P_0 + \frac{3}{2}\rho g R) \pi R^2$. Let's try to express our force in the form $(P_0 + x\rho g R) y\pi R^2$. $F = \pi R^2 (\frac{3}{4}P_0 + \frac{3}{2}\rho g R)$. If we try to force $y$ to be an integer, let's consider the $\rho g R$ term. $yx = 3/2$. If $y=1$, $x=3/2$. $n = 13/(3/2) = 26/3$. If $y=2$, $x=3/4$. $n = 13/(3/4) = 52/3$. If $y=3$, $x=1/2$. $n = 13/(1/2) = 26$. Let's check if $y=3$ and $x=1/2$ is consistent with the $P_0$ term. If $y=3$, the $P_0$ term in the given formula is $y P_0 \pi R^2 = 3 P_0 \pi R^2$. Our calculated $P_0$ term is $\frac{3}{4}P_0 \pi R^2$. This does not match.

Let's consider the case where the question implies the force is $F = (P_0 + x\rho g \times \text{some length}) \times (\text{Area})$. And the given form is $(P_0 + x\rho gr)y\pi R^2$. Let's assume $r=R$. $F = y P_0 \pi R^2 + yx \rho g R \pi R^2$. Our calculated force is $F = \frac{3}{4}P_0 \pi R^2 + \frac{3}{2}\rho g R \pi R^2$. If we assume that the coefficient of $\rho g R$ is $3/2$, and $y$ is an integer. Let $y=3$. Then $3x = 3/2$, so $x=1/2$. If $x=13/n$, then $1/2 = 13/n$, so $n=26$. Let's check the $P_0$ term. If $y=3$, the $P_0$ term is $3 P_0 \pi R^2$. Our calculated $P_0$ term is $\frac{3}{4}P_0 \pi R^2$. This does not match.

There seems to be an inconsistency in the problem statement or the given formula. However, if we assume that the $\rho g R$ term is the one that should be matched, and $y$ is an integer, the most plausible solution arises from $y=3$ and $x=1/2$. This gives $n=26$. Let's check if there's an interpretation where the $P_0$ term also aligns. If $y=3$, the given formula's $P_0$ term is $3 P_0 \pi R^2$. Our calculated force has $P_0$ term $\frac{3}{4}P_0 \pi R^2$. The ratio is $3 / (3/4) = 4$. This suggests that the area factor might be different.

Let's assume the given formula is $(P_0 + x\rho gr) \times A_{given}$. And $A_{given} = y\pi R^2$. Our calculated force is $F = (P_0 + 2\rho g R) \times (\frac{3}{4}\pi R^2)$. If we assume $x=2$ and $r=R$. Then $F = (P_0 + 2\rho g R) \times (\frac{3}{4}\pi R^2)$. And the given form is $(P_0 + 2\rho g R) \times y\pi R^2$. This implies $y = 3/4$.

Let's assume the problem meant that the force is $F = (P_0 + x\rho g R) \times (\text{Area})$. And the Area is given by $y\pi R^2$. So $F = (P_0 + x\rho g R) \times (y\pi R^2)$. We calculated $F = (P_0 + 2\rho g R) \times (\frac{3}{4}\pi R^2)$. For these to match, we need $x=2$ and $y=3/4$.

Given $y$ is an integer, let's re-examine the coefficients. $F = \frac{3}{4}P_0 \pi R^2 + \frac{3}{2}\rho g R \pi R^2$. Given $F = y P_0 \pi R^2 + yx \rho g R \pi R^2$. If we assume that the problem meant that the force is expressed in a form where $y$ is an integer, and the ratio of the coefficients of $P_0$ and $\rho g R$ is preserved. Let the actual force be $F_{actual} = (\frac{3}{4}P_0 + \frac{3}{2}\rho g R) \pi R^2$. Let the given form be $F_{given} = y (P_0 + x\rho g R) \pi R^2$. If $y$ is an integer, and $x=13/n$. If we consider the $\rho g R$ term: $yx = 3/2$. If we consider the $P_0$ term: $y = 3/4$. This is the contradiction.

Let's assume the problem setter intended that the $\rho g R$ term is the key. $yx = 3/2$. If we assume $y=3$ (an integer), then $3x = 3/2 \implies x = 1/2$. Given $x = 13/n$, we have $1/2 = 13/n$. $n = 13 \times 2 = 26$. Let's check if this is consistent with the $P_0$ term. If $y=3$, the $P_0$ term in the given formula is $3 P_0 \pi R^2$. Our calculated $P_0$ term is $\frac{3}{4}P_0 \pi R^2$. The ratio of coefficients is $3 / (3/4) = 4$. This implies that the area factor $y$ in the given formula is not just multiplying the pressure term, but is part of a larger structure.

Let's go back to $F = (P_0 + x\rho gr)y\pi R^2$. And $F = \frac{3}{4}P_0 \pi R^2 + \frac{3}{2}\rho g R \pi R^2$. Let $r=R$. $y P_0 \pi R^2 + yx \rho g R \pi R^2 = \frac{3}{4}P_0 \pi R^2 + \frac{3}{2}\rho g R \pi R^2$. If we assume that the problem implies that the ratio of the coefficients of $P_0$ and $\rho g R$ in the given formula should match the ratio in our calculated force, and $y$ is an integer. Ratio of coefficients of $P_0$ and $\rho g R$ in calculated force: $(\frac{3}{4}) / (\frac{3}{2}) = \frac{3}{4} \times \frac{2}{3} = \frac{1}{2}$. Ratio of coefficients in given formula: $y / (yx) = 1/x$. So, $1/x = 1/2 \implies x=2$. If $x=2$, then $2 = 13/n$, so $n = 13/2$. This is not an integer.

Let's assume the given formula should be interpreted as $F = (P_0 + x\rho g r) \times A_{given}$. And $A_{given} = y\pi R^2$. Our calculated force is $F = (P_0 + 2\rho g R) \times (\frac{3}{4}\pi R^2)$. If we assume $x=2$ and $r=R$. Then $F = (P_0 + 2\rho g R) \times (\frac{3}{4}\pi R^2)$. And the given form is $(P_0 + 2\rho g R) \times y\pi R^2$. This implies $y = 3/4$.

Let's assume that the problem setter intended for $y$ to be an integer such that when multiplied by the pressure term, and then by some area, it matches. Given the constraint that $y$ is an integer, and $x=13/n$. Let's assume that the $\rho g R$ term is the determining factor for $x$ and $y$. $yx = 3/2$. If we assume $y=3$ (integer), then $3x = 3/2 \implies x = 1/2$. Given $x = 13/n$, we have $1/2 = 13/n$. $n = 13 \times 2 = 26$.

This interpretation assumes that the $P_0$ term in the given formula is somehow scaled to match the calculated force. Let's verify this assumption. If $y=3$ and $x=1/2$, the given formula is $(P_0 + \frac{1}{2}\rho g R)3\pi R^2 = 3P_0 \pi R^2 + \frac{3}{2}\rho g R \pi R^2$. Our calculated force is $\frac{3}{4}P_0 \pi R^2 + \frac{3}{2}\rho g R \pi R^2$. The $\rho g R$ terms match. The $P_0$ terms do not match. However, if we assume the problem meant that the force is $(P_0 \times \text{something} + x\rho g r \times \text{something else}) \times y\pi R^2$. Or if the $y$ factor is meant to scale the entire expression. Let's assume the structure $(P_0 + x\rho gr)y\pi R^2$ implies that the $\rho g r$ term is the one that should be matched. Given $yx = 3/2$. And $y$ is an integer. The smallest integer $y$ that can give a rational $x$ such that $x=13/n$ leads to an integer $n$ is to make $y$ a divisor of $3/2$. Let's try integer values for $y$. If $y=1$, $x=3/2$. $n=13/(3/2)=26/3$. If $y=2$, $x=3/4$. $n=13/(3/4)=52/3$. If $y=3$, $x=1/2$. $n=13/(1/2)=26$. If $y=4$, $x=3/8$. $n=13/(3/8)=104/3$.

The value $n=26$ is obtained when $y=3$ and $x=1/2$. This is the most plausible solution given the constraints, assuming a potential discrepancy in the $P_0$ term matching. The question is likely designed such that the $\rho g R$ term is the primary focus for determining $x$ and $y$.

Final check: Calculated force: $F = \frac{3}{4}P_0 \pi R^2 + \frac{3}{2}\rho g R \pi R^2$. Given form: $(P_0 + x\rho gr)y\pi R^2$. Assume $r=R$. Given form: $y P_0 \pi R^2 + yx \rho g R \pi R^2$. We found $y=3$, $x=1/2$. This gives $3 P_0 \pi R^2 + \frac{3}{2}\rho g R \pi R^2$. The $\rho g R$ term matches. The $P_0$ term does not. However, if we consider the possibility that the question implies the force is $F = (\text{scaled } P_0 \text{ term} + \rho g \text{ term})$ and the $y$ multiplier is applied to the entire expression. If $F = y \times (\text{Pressure}) \times (\text{Area})$. Let Area $= \pi R^2$. $F = y (P_0 + x\rho g R) \pi R^2 = y P_0 \pi R^2 + yx \rho g R \pi R^2$. This is the same equation as before.

Given $x = 13/n$ and $n$ is likely an integer. If $x=1/2$, then $1/2 = 13/n \implies n=26$. This is the most consistent integer value for $n$.