Explore topic-wise InterviewSolutions in Current Affairs.

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1.

Let M be the mass and L be the length of a thin uniform rod. In first case, axis of rotation is passing through centre and perpendicular to the length of the rod. In second case, axis of rotation is passing through one end and perpendicular to the length of the rod. The ratio of radius of gyration in first case to second case isA. 1B. `1/2`C. `1/4`D. `1/8`

Answer» Correct Answer - B
According to question, Moment of inertia of rod whose axis is passing through centre and perpendicular to the rod is given by `l=(ML^2)/12` …(i)
in terms of radius of gyration `l=MK^2` …(ii)
Comparing Eqs. (i) and (ii) , we get `MK_1^2=(ML^2)/12`
Rightarrow `K_1=L/(2sqrt3`
In second case, moment of inertia when axis is passing through one of the end is given by `l=(ML^2)/3`
Similarly in terms of radius of gyration `l= M_2^2`
From Eqs. (iv) and (v), we get `(ML^2)/3=MK_2^2`
`K_2=L/sqrt3`
Again taking the ratio of K, and K, from Eqs.(iv) and (vi),
We have, `K_1/K_2=(Lxxsqrt3)/(2sqrt3xxL)=1/2`
2.

If A and B are foot of perpendicular drawn from point Q(a,b,c) to the planes yz and zx, then equation of plane through the point A,B, and O isA. `(x)/(a)+(y)/(b)-(z)/(c)=0`B. `(x)/(a)-(y)/(b)+(z)/(c)=0`C. `(x)/(a)-(y)/(b)-(z)/(c)=0`D. `(x)/(a)+(y)/(b)+(z)/(c)=0`

Answer» Correct Answer - A
The foot of perpendicular from point Q(a,b,c) to the yz plane is A(0,b,c) and the foot of perpendicular from point Q to the foot of perpendicular from point Q to zx plane in B(a, 0, c).
Let the.equation of plane passing through the point (0, 0, 0) be
Ax+By+Cz=0 ... (i)
Also it is paring through the point A( 0, b, c) and B(a,0,c)
`therefore 0+Bc+C c=0`
`and Aa+0+C c=0`
`Rightarrow C c=Bb`
and `C c=-A a `
`therefore A=-(k)/(a), B=-(k)/(b)`
and `C=(k)/(c)`
From Eq. (i) , `(-k)/(a)x-(k)/(b)y+(k)/(c)z=0`
`Rightarrow -(x)/(a)-(y)/(b)+(z)/(c)=0 or (x)/(a)+(y)/(b)+(z)/(c)=0`