A cubical block of wood of edge 3cm floats in water. The lower surface of the cube just touches the free end of a vertical spring fixed at the bottom of the pot. The maximum weight that can be put on the block without wetting it is (P) Newton.
Then the value of 2P is (Spring constant = 50 N/m, density of wood $= 800 k g / m^{3}, g = 10 m / s^{2}$ )

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let a =edge of the block =3cms= height of the block above the water surface in floating condition.
From F.B.D., we have
$ρ a^{3} g = ρ_{w} a^{2} (a - x) g ... (1)$
$ρ a^{3} g + W = k x + ρ_{w} a^{3} g ... (2)$
From eqn. (1), we get
$x = a (1 - \frac{ρ}{ρ_{w}}) = 3 (1 - 0.8) = 0.6 c m$
From eqn. (2). we get
$W = k x + (ρ_{w} - ρ) a^{3} g$
$= 50 \times 0.006 + (1000 - 800) \times {(3 \times 10^{- 2})}^{3} \times 10 = 0.354 = (20)$ Newtons