Moment of inertia of disc of radius R about a diametric axis is 25kg/m². MI of disc about a parallel axis at a distance R/2 from centre is

Moment of inertia of disc of radius R about a diametric axis is 25kg/m

1.	Moment of inertia of disc of radius R about a diametric axis is 25kg/m². MI of disc about a parallel axis at a distance R/2 from centre is
Answer» 50Explanation:MI about diameter = M(R^2)/4 =25 kg/m^2 MI about ll(L) axis=M(R^2)/4 + M(R^2)/2 = 2M(R^2)/4 = 225 = 50 kg/m^2

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Guys...this may seems to be hard...bt no...it's not...I need to know the solution of only part:-(d) & (e)(a)The moment of inertia of a disc of radius 0.5 m about its geometric axis is 2kg-m2. If a string is tied toits circumference and a force of 10 Newton is applied, the value of torque with respect to this axis will be(1) 2.5 N-m21275 N-m(3) 10 N-m(4) 20 N-m(b)In the above question, if the disc executes rotatory motion, its angular acceleration will be :-(1) 2.5 rad/sec(2) 5 rad/sec(3) 10 rad/sec(4) 20 rad/sec(c)In the above question, the value of its angular velocity after 2 seconds will be :-(1) 2.5 rad/sec(2) 5 rad/sec(3) 10 rad/sec(4) 20 rad/sec(d)In the above question, the change in angular momentum of disc in first 2 seconds in Nm second will be(1) 2.5(2) 5(3) 10(4) 20(e)In the above question, angular displacement of the disc, in first two second will be in radian:-(1) 2.5(2) 5(3) 10(4) 20Plz guys...I just don't wanna copy it in my hw...I wanna learn how the correct answer came out...so plz explain your solution with proper formula and sign and don't spam...plzWould be very thankful to you...!!!
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Moment of inertia of disc of radius R about a diametric axis is 25kg/m². MI of disc about a parallel axis at a distance R/2 from centre is

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