Match List – I with List – II :

List – I                                                                               List – II

(a) h (Planck’s constant)                                               (i) $\left[ML{T}^{-1}\right]$  

(b) E (kinetic energy)                                                     (ii) $\left[M{L}^2{T}^{-1}\right]$        

(c) V (electric potential)                                                (iii) $\left[M{L}^2{T}^{-2}\right]$  

(d) P (linear momentum)                                             (iv) $\left[M{L}^2{I}^{-1}{T}^{-3}\right]$  

Choose the correct answer from the options given below:

h = u²/2g, u = √2gh
Now, S = h/3
S = ut + ½at²
h/3 = √2ght - ½gt²
t² - 2√ (2h/g)t + 2h/3g = 0
Using quadratic formula for t:
t = ( 2√ (2h/g) ± √ (8h/g) - 4 (2h/3g) / 2
t = √ (2h/g) ± √ (2h/g - 2h/3g) = √ (2h/g) ± √ (4h/3g)
t? /t? = (√ (2h/g) - √ (4h/3g) / (√ (2h/g) + √ (4h/3g)
t? /t? = (√2 - √ (4/3) / (√2 + √ (4/3) = (√6 - 2)/ (√6 + 2)
(Note: There is a calculation error in the provided solution. Re-evaluating the physics.)
h/3 = (√2gh)t - ½gt²
(g/2)t² - (√2gh)t + h/3 = 0
t = (√2gh ± √ (2gh - 4 (g/2) (h/3)/g = (√2gh ± √ (4gh/3)/g
t? /t? = (√2gh - 2√gh/√3)/ (√2gh + 2√gh/√3) = (√6 - 2)/ (√6 + 2)
The answer in the options suggests a different initial equation.
Let's re-examine the provided solution:
0 = t² - √ (8h/g)t + 2h/3g
This implies the initial equation was: ½gt² - (√2gh)t + h/3 = 0 which is correct.
The step from here to t? /t? is incorrect in the image.