Question

Three blocks of different masses connected with inextensible string are pulled by a force \(F\) on a frictionless surface as shown in the figure. The ratio of tensions \(T_1\) to \(T_2\) is

• A. \(1:5\)
• B. \(5:1\)
• C. \(10:1\)
• D. \(1:10\)

Hint:

Find the common acceleration, then apply Newton second law to the blocks behind each string.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The figure shows three blocks of 2 kg, 8 kg and 12 kg on a frictionless surface, joined by strings and pulled by a force \(F\) on the 12 kg block. The string \(T_2\) joins 2 kg and 8 kg, and \(T_1\) joins 8 kg and 12 kg.

Step 2: Common acceleration
All blocks move together:
\[ a=\frac{F}{2+8+12}=\frac{F}{22} \]

Step 3: Tension \(T_2\)
It pulls only the 2 kg block:
\[ T_2=2a=\frac{2F}{22} \]

Step 4: Tension \(T_1\)
It pulls the 2 kg and 8 kg blocks together (10 kg):
\[ T_1=10a=\frac{10F}{22} \]

Step 5: Ratio
\[ \frac{T_1}{T_2}=\frac{10}{2}=5 \]
So \(T_1:T_2=5:1\), option (B).

Final Answer:
The tension nearer the force carries 10 kg and the other carries 2 kg, so the ratio is 5 to 1, option (B). \[ \boxed{5:1} \]