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[" A rod of length "6m" has specific gra...

[" A rod of length "6m" has specific gravity "rho(=25/36)." One end of the rod is tied to a "5m" long "],[" rope,which in turn is tied to the floor of a pool "10m" deep,as shown.Find the length (in "m)" of "],[" the part of rod which is out of water "],[qquad [1],[10m],[" Key ":0001.00]],[" Key ":0001.00],[" Ley ":0001.00" ."]

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A rod of length 6 m has specific gravity rho (= 25//36) . One end of the rod is tied to a 5 m long rope, which in turn is tied to the floor of a pool 10 m deep, as shown. Find the length (in m ) of the part of rod which is out of water.

A rod of length 6 m has specific gravity rho (= 25//36) . One end of the rod is tied to a 5 m long rope, which in turn is tied to the floor of a pool 10 m deep, as shown. Find the length (in m ) of the part of rod which is out of water.

A rod of length 2 m is held vertically with one of its end on the floor. It is then allowed to fall. Calculate the speed of the other end when it strikes the floor. Assume that the end of the rod which is on the floor does not slip.

A uniform rod of length 1.00 m is suspended through an end and is set into oscillation with small amplitude under gravity. Find the time period of oscillation.

A uniform rod of length 1.00 m is suspended through an end and is set into oscillation with small amplitude under gravity. Find the time period of oscillation.

A uniform rod of length 1.00 m is suspended through an end and is set into oscillation with small amplitude under gravity. Find the time period of oscillation.

The length of the longest rod which can be laid across a floor of a rectangular room 12 m in length and 5 m in breadth will be:

A uniform rod of length 2.0 m specific gravity 0.5 and mass 2 kg is hinged at one end to the bottom of a tank of water (specific gravity = 10) filled upto a height o f 1.0 m as shown in figure. Taking the case theta =- 0^(@) the force exerted by the hings on the rod is (g = 10 m//s^(2))

A uniform rod of length 2.0 m specific gravity 0.5 and mass 2 kg is hinged at one end to the bottom of a tank of water (specific gravity = 10) filled upto a height o f 1.0 m as shown in figure. Taking the case theta =- 0^(@) the force exerted by the hings on the rod is (g = 10 m//s^(2))