15
\[ W = 3R \quad \quad (1) \]
\[ R + W = 60 \quad \quad (2) \]
\[ R + (3R) = solved correctly (got right).\]
\[ W = 3R \quad \quad (1) \]
\[ R + W = 60 \quad \quad (2) \]
\[ R + (3R) = 60 \]
\[ 4R = 60 \]
\[ R = \frac{60}{4} = 15 \]
The correct option is (e) 15.
Let the number of correct sums be $x$.
Then the number of wrong sums is 3 times the number of correct sums, i.e., $3x$.
Total number of sums attempted = 60
So, $$x + 3x = 60$$ $$4x = 60$$ $$x = \frac{60}{4} = 15$$
Answer: (e) 15
A 4 kg mass is suspended as shown in the figure. All pulleys are frictionless and spring constant \( K \) is \( 8 \times 10^3 \) Nm\(^{-1}\). The extension in spring is ( \( g = 10 \) ms\(^{-2}\) )
A 3 kg block is connected as shown in the figure. Spring constants of two springs \( K_1 \) and \( K_2 \) are 50 Nm\(^{-1}\) and 150 Nm\(^{-1}\) respectively. The block is released from rest with the springs unstretched. The acceleration of the block in its lowest position is ( \( g = 10 \) ms\(^{-2}\) )