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हम आपके सपनो को साकार करेंगे III
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The moment of inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through. 2012 B C D A
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In a rectangle ABCD (BC=2AB). The moment of inertia is minimum anout which the axis will be 1993 BC BD HF EG
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ABC is a traiangular plate of uniform thickness. The sides are in the ratio shown in the figure. "I" _"AB" , "I" _"BC" and "I" _"CA" are the moments of inertia of the plate about AB, BC and CA respectively. Which one of the following relations is correct? 1995 "I" _"AB" > "I" _"BC" "I" _"BC" > "I" _"AB" "I" _"AB" + "I" _"BC" = "I" _"CA" "I" _"CA" IS MAXIMUM
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There is a flat uniform triangular plate ABC, sides AB= 4 cm, BC= 3 cm and ∠ABC=90∘. The moment of inertia of the plate about AB, BC and CA as axis are I1, I2 and I3 respectively. Which one of the following is true? 𝐼_3> 𝐼_2 𝐼_2> 𝐼_1 𝐼_3> 𝐼_1 𝐼_1> 𝐼_2
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A composite disc to be made using equal masses of aluminimum and iron so that it has as high a moment of inertia as possible. This possible when. 2002 The surfaces of the discs are made of iron with aluminium inside The whole of aluminum is kept in the core and the iron at the outer rim of the disc The whole of the iron is kept in the core and the aluminium at the outer rim of the disc The whole disc is made with thin alternate sheets to iron and aluminimum.
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Three particles. each of mass m gram, are situated at the vertices of an equilateral triangle ABC of side l cm (as shown in the figure). The moment of inertia of the system about a line AX perpendicular to AB and in the plane of ABC, in gram-cm2 units will be :2004 2/3 〖𝑀𝐿〗^2 3/4 〖𝑀𝐿〗^2 2〖𝑀𝐿〗^2 5/4 〖𝑀𝐿〗^2
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A thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is 90°. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is : 2008 (1) ML2/6 (2) √2ML2/24 (3) ML2/24 (4) ML2/12
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Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ω0 is minimum, is given by :2015 𝑚_1/𝑚_2 L 𝑚_2/𝑚_1 L 𝑚_2/(𝑚_1+𝑚_2 ) L 𝑚_1/(𝑚_1+𝑚_2 ) L
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From a circular ring of mass 'M' and radius 'R' an arc corresponding to a 90° sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is 'K' times 'MR2'. Then the value of 'K' is : 2021 ncert page 163 𝟏/𝟖 𝟑/𝟒 𝟕/𝟖 𝟏/𝟒
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