A wooden block of mass 1 kg and density 800 kg m- 3 is held stationary, with the help of a string, in a container filled with water of density 1000 kg m- 3, as shown in the figure. If the container is moved upwards with an acceleration of 2 ms- 2 , then the tension in the string will be ( take g=10 ms- 2 )

Question

A wooden block of mass 1 kg and density 800 kg m- 3 is held stationary, with the help of a string, in a container filled with water of density 1000 kg m- 3, as shown in the figure. If the container is moved upwards with an acceleration of 2 ms- 2 , then the tension in the string will be ( take g=10 ms- 2 ) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-kmc5yi4l1rqrhhad.jpg" /> (A) 2N (B) 3 N (C) 5N (D) 6N

Tardigrade Admin · Accepted Answer

Wâ²=m(.g+a.)=1× 12=12 N B prime=ρl × v ×(g+a)=1000 × (1/800) ×(12)=15 N ∴ T=Bâ²-Wâ²=15-12=3N

$2 N$

$3 N$

$5 N$

$6 N$

$W^{'} = m (g + a) = 1 \times 12 = 12 N$
$B^{'} = ρ_{l} \times v \times (g + a) = 1000 \times \frac{1}{800} \times (12) = 15 N$
$∴ T = B^{'} - W^{'} = 15 - 12 = 3 N$

2N

3N

5N

6N

W′=m(g+a)=1×12=12N B′=ρl​×v×(g+a)=1000×8001​×(12)=15N ∴T=B′−W′=15−12=3N

$2 N$

$3 N$

$5 N$

$6 N$

$W^{'} = m (g + a) = 1 \times 12 = 12 N$
$B^{'} = ρ_{l} \times v \times (g + a) = 1000 \times \frac{1}{800} \times (12) = 15 N$
$∴ T = B^{'} - W^{'} = 15 - 12 = 3 N$