Two spheres each of volume 250 cc but of relative densities 0.8 and 1.2 are connected by a string and the combination is immersed in a liquid in vertical position as shown in the figure. The tension in the string is (k/18) N. The value of k is.(Given, g=10 ms -2 )

Question

Two spheres each of volume 250 cc but of relative densities 0.8 and 1.2 are connected by a string and the combination is immersed in a liquid in vertical position as shown in the figure. The tension in the string is (k/18) N. The value of k is.(Given, g=10 ms -2 ) (A)  (B)  (C)  (D)

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0fbaaee926bdc47f3135229a500c3223-.png" /> T+m g=BF1 Or T+(0.8)(250) g=(250) ρL g ...(i) T+BF2=m2 g Or T+(250)(ρL)(g)=(1.2)(250)(g) ...(ii) From equations (i) and (ii), T=0.5 N ∴ (k/18)=(1/2) ⇒ k=9

Q. Two spheres each of volume $250 cc$ but of relative densities $0.8$ and $1.2$ are connected by a string and the combination is immersed in a liquid in vertical position as shown in the figure. The tension in the string is $\frac{k}{18} N$ . The value of $k$ is_____.(Given, $g = 10 m s^{- 2}$ )

$T + m g = B_{F_{1}}$
Or $T + (0.8) (250) g = (250) ρ_{L} g$ ...(i)
$T + B_{F_{2}} = m_{2} g$
Or $T + (250) (ρ_{L}) (g) = (1.2) (250) (g)$ ...(ii)
From equations (i) and (ii), $T = 0.5 N$
$∴ \frac{k}{18} = \frac{1}{2}$
$\Rightarrow k = 9$

Q. Two spheres each of volume 250cc but of relative densities 0.8 and 1.2 are connected by a string and the combination is immersed in a liquid in vertical position as shown in the figure. The tension in the string is 18k​N. The value of k is_____.(Given, g=10ms−2 )

T+mg=BF1​​ Or T+(0.8)(250)g=(250)ρL​g ...(i) T+BF2​​=m2​g Or T+(250)(ρL​)(g)=(1.2)(250)(g) ...(ii) From equations (i) and (ii), T=0.5N ∴18k​=21​ ⇒k=9

Q. Two spheres each of volume $250 cc$ but of relative densities $0.8$ and $1.2$ are connected by a string and the combination is immersed in a liquid in vertical position as shown in the figure. The tension in the string is $\frac{k}{18} N$ . The value of $k$ is_____.(Given, $g = 10 m s^{- 2}$ )

$T + m g = B_{F_{1}}$
Or $T + (0.8) (250) g = (250) ρ_{L} g$ ...(i)
$T + B_{F_{2}} = m_{2} g$
Or $T + (250) (ρ_{L}) (g) = (1.2) (250) (g)$ ...(ii)
From equations (i) and (ii), $T = 0.5 N$
$∴ \frac{k}{18} = \frac{1}{2}$
$\Rightarrow k = 9$