Let θ1,l1 be the angle of inclination and distance travelled from top to bottom respectively on plane (1) and θ2,l2 be the angle of inclination and distance travelled from top to bottom respectively on plane (2) From diagram
θ1>θ2 i.e. sin‌θ1>sin‌θ2 or

sin‌θ1

sin‌θ2

>1...(i)
h= height of each inclined plane =l1‌sin‌θ1=l2‌sin‌θ2
(a) Yes. At the top of the planes, the sphere has only P.E = mgh
where m = mass of the sphere
When the sphere rolls down from the top to the bottom, converted into K.E.
∴ If v1‌and‌v2 be its linear speeds at the bottom of the planes, i.e. (1) and (2) respectively, then
=(

1

2

mv12+

1

2

mk2

v12

R2

) or 2gh=v12(1+

k2

R2

) and 2gh=v22 (1+

k2

R2

)
∴v12=

2gh

(1+ k2 R2 )

...(ii)
and v22=

2gh

(1+ k2 R2 )

...(iii)
where I = MI of the sphere, ω = its angular speed, k is the radius of gyration
From (ii) and (iii) it is clear that the sphere reaches the bottom with same speed in each case.
(b) To find the time of rolling motion : Yes, it will take longer time down one plane than the other. It will be longer for the plane having smaller angle of inclination.
(c) Explanation : Let t1‌and‌t2 be the time taken by the sphere in rolling on plane (1) and (2) respectively.
Acceleration of solid sphere on an inclined plane is given by,
a=

g‌sin‌θ

(1+ k2 R2 )

Now for solid sphere 1+

k2

R2

=1+

2 5 R2

R2

=

7

5

If a1‌and‌a2 be the accelerations of the sphere on inclined plane (1) and (2) respectively, then
a1=

g‌sin‌θ1

7 5

=

5

7

g‌sin‌θ1
Similarly a2=

5

7

g‌sin‌θ2
S=ut+

1

2

at2, we get t12=

2l1

a1

=

2h∕sin‌θ1

5 7 g‌sin‌θ1

=

14h

5g(sin‌θ1)2

... (iv)
and t22=

2l2

a2

=

2h∕sin‌θ2

5 7 g‌sin‌θ2

=

14h

5g(sin‌θ2)2

...(v)
It gives

t12

t22

=

(sin‌θ2‌ )2

(sin‌θ1)2

⇒

t1

t2

=

sin‌θ2

sin‌θ1

...(vi)
Now from (i)

sin‌θ1

sin‌θ2

>1 or

sin‌θ2

sin‌θ1

<1
∴(

sin‌θ2

sin‌θ1

)2<1...(vii)
∴ From (vi) and (vii), we get

t1

t2

<1 or $t_{1}It is due to the reason that a∝sin‌θ and t is inversely proportional to ‘a’ or sin‌θ.