Question

1. a simple barometer is constructed using a liquid q1 of density 16.2cm³. calculate the least length of the tube to be used to measure pressure of 1.5atm. density of mercury = 13.6cm³, 1atm = 76.0 cm of mercury a. 1358mm b. 957mm c. 760mm d. 638mm 28. f₁=20n f₂=20n 0cm a b 50cm a bar ab is balanced horizontally on two knife edges as illustrated in the diagram above. determine the weight of the bar.

Explanation:

Question 27

Step 1: Recall Pressure Formula

Pressure \( P =
ho gh \), where \(
ho \) is density, \( g \) is acceleration due to gravity, \( h \) is height. For same pressure, \(
ho_1 h_1=
ho_2 h_2 \). Here, pressure from mercury (1 atm) and liquid Q1 (1.5 atm) relate. First, find height of mercury for 1.5 atm: \( h_{Hg}=1.5\times76.0\space cm \). Then equate pressures: \(
ho_{Hg} h_{Hg}=
ho_{Q1} h_{Q1} \).

Step 2: Calculate Height of Mercury for 1.5 atm

\( h_{Hg}=1.5\times76.0 = 114\space cm \)

Step 3: Solve for \( h_{Q1} \)

Using \(
ho_{Hg} h_{Hg}=
ho_{Q1} h_{Q1} \), so \( h_{Q1}=\frac{
ho_{Hg} h_{Hg}}{
ho_{Q1}} \). Substitute \(
ho_{Hg}=13.6\space g/cm^3 \), \( h_{Hg}=114\space cm \), \(
ho_{Q1}=16.2\space g/cm^3 \) (assuming density unit is \( g/cm^3 \), typo in question).
\( h_{Q1}=\frac{13.6\times114}{16.2} \space cm \)
\( h_{Q1}=\frac{1550.4}{16.2}\approx95.7\space cm = 957\space mm \)

Step 1: Identify Torque Equilibrium

For horizontal equilibrium, sum of clockwise torques = sum of counter - clockwise torques. Let weight of bar be \( W \), acting at its center (since it's uniform, center at 25 cm from A and B? Wait, length of bar is 50 cm (from 0cm to 50cm). So center is at 25 cm. Let's take moments about A.

Step 2: Calculate Torques

Force \( F_1 = 20N \) at A (distance 0 from A, torque \( \tau_1 = 20\times0 = 0 \)). Force \( F_2 = 20N \) at B (distance 50 cm from A, torque \( \tau_2=20\times50 \) (counter - clockwise? Wait, no: the weight \( W \) acts downward at center (25 cm from A), creating clockwise torque about A. \( F_2 \) acts upward at B, creating counter - clockwise torque about A. Wait, correct approach: Let's take moments about A. The weight \( W \) acts at the mid - point (25 cm from A) downward, so torque due to \( W \) is \( W\times25 \) (clockwise). Torque due to \( F_2 \) is \( F_2\times50 \) (counter - clockwise). Since the bar is in equilibrium, clockwise torque = counter - clockwise torque. Wait, no: actually, the forces \( F_1 \) and \( F_2 \) are upward, weight \( W \) is downward. Let's use force equilibrium first: \( F_1+F_2 = W \)? No, wait, \( F_1 = 20N \), \( F_2 = 20N \), upward forces. Downward force is \( W \). So \( F_1 + F_2=W \)? But that would be \( W = 40N \), but that's wrong. Wait, no, the bar is balanced on two knife edges, so we need to take moments. Let's take moment about A: the torque due to \( W \) (acting at center, 25 cm from A) is \( W\times25 \) (clockwise), torque due to \( F_2 \) (acting at 50 cm from A) is \( F_2\times50 \) (counter - clockwise). Torque due to \( F_1 \) is zero (distance from A is 0). So \( W\times25=F_2\times50 \).

Step 3: Solve for \( W \)

\( W\times25 = 20\times50 \)
\( W=\frac{20\times50}{25}=40N \)
Wait, but let's check with moment about B. Torque due to \( W \) (distance from B is 25 cm) is \( W\times25 \) (counter - clockwise), torque due to \( F_1 \) (distance from B is 50 cm) is \( F_1\times50 \) (clockwise). So \( W\times25=F_1\times50 \), \( W=\frac{20\times50}{25}=40N \). Also, force equilibrium: \( F_1 + F_2=W \), \( 20 + 20 = 40 \), which matches.

Answer:

B. 957mm

Question 28