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It is not possible to get admitted to the Bachelor of Physiotherapy (BPT) course Topiwala National Medical College (TNMC), Mumbai through MH CET and without NEET to the 2025-2026 session.

Here's why:

BPT at The New Medical College Mumbai: NEET Is Now Compulsory
Since the latest rules, NEET-UG is now compulsory to possess the admission to top physiotherapy colleges in Maharashtra (such as Topiwala National Medical College, Mumbai) since the 2025-26 study year.

TNMC offers admission to BPT course, which follows an application and counselling procedure that is channelized by the Directorate of Medical Education and Research (DMER), Maharashtra, based on NEET-UG scores in the short-listing of candidates.

In 2025, MH CET Is Inadequate to get admission to BPT at TNMC?
In the past, a number of colleges (both, private and government) in Maharashtra used to admit on the basis of MH CET in BPT. But this has changed as rules with regard to the current/ next admission cycles now entail NEET to secure admissions in BPT courses in most of the government colleges, including TNMC.

In the 2025-26 batch, MH CET will not alone give BPT at TNMC; NEET-UG will also be necessary.

General Eligibility
You have a passing percentage (60%) in 12th which is good in terms of marks (normally minimum required percentage is 50 in PCB).

Which, however, NEET-UG qualification is a must along with 12th (PCB) to apply to BPT in TNMC Mumbai.

Other Maharashtra State Entrance
It is possible that some of the privately run colleges in Maharashtra accept the 12th board results or MH CET to get the BPT seats, but the top government colleges (such as TNMC) now accept NEET-UG.

In short:
BPD aspirants would need to pass NEET-UG in order to have a chance in BPT at Topiwala National Medical College in 2025. The marks of MH CET and 12 th are not the only factors which determine the admission to BPT at this college in the current admission cycle

3.14

The distance from home to market = 2.5 km = 2500 m

The speed of the walking while going = 5 kmph = 1.388 m/s

The speed of walking back from market to home = 7.5 kmph = 2.083 m/s

(a) Magnitude of the average velocity = Displacement / time = 0, since the total displacement is zero.

(b)

i. Time taken to reach the market = distance / onward speed = 2500/1.388 = 1801 s = 30 minutes. So the average speed over 0 – 30 min is 5 kmph

ii. Time taken to reach back home = 2500 / 2.083 = 20 minutes. So the average speed = total distance covered / total time taken = (2500 +2500)/ (30 +20) m/min = 100 m/min = 6 kmph

iii. Average speed over the interval of time 0 to 40 mins = Distance covered / time taken = 2500 + (2.083 * 10 * 600) / (40 * 60) = 1.5624 m/s = 5.62 kmph

3.13

(a) Let us take the example of football world cup 2022. During Argentina match, Di Mario (A) passed the ball to Messi (B). Messi instantly passed the ball to Di Mario. Now the magnitude of displacement of the ball is 0 as the ball returns to Di Mario at the initial position. But the total length covered by the ball is AB + BA = 2AB. Hence total length covered by the particle (the ball) is more than the magnitude.

(b) If t is the time taken to cover the entire distance, the magnitude of the average velocity of the ball over time interval t = Magnitude of displacement / t = 0 / t = 0

The average speed of the ball = Total length of the path / time interval = 2AB / t

Thus the second quantity is greater than the first.

If the ball moves only in one direction from A to B, both the quantities are equal (considering one-dimensional motion)

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3.10

(a) The direction of the acceleration is downward during the upward motion of the ball because the acceleration is due to gravity. Gravity always pulls the object toward the centre of the earth. 

(b) Velocity: At the peak the velocity is zero because the ball momentarily stops before falling downwards. 

(c) If we consider the highest point x = 0 m, t= 0 s,

During upward motion of the ball before it reaches the maximum height, x = +ve, velocity = -ve, acceleration = +ve. During downward motion, x = +ve, velocity = +ve, acceleration = +ve

(d) At the highest point, final velocity v = 0, initial velocity = 29.4 m/s, acceleration = 9.8m/s²

From the equation v2-u2 = 2gs, we get s = (29.4)²/2 * 9.8 = 44.1 m. From v –u = gt we get t = 29.4/9.8 s = 3 s. So the total time taken by the ball to reach player's hand = 3 * 2s = 6 s

3.9 Cycling speed $V_c$ = 20 kmph

Let as assume speed of the bus is $V_b$

The relative velocity of the buses in the direction of motion of the cyclist = $V_b$ - $V_c$

Buses go past him every 18 mins = 18/60h = 0.3h

Distance covered = ( $V_b-V_c$ ) * 0.3

Since the buses leave every T minutes, ( $V_b-V_c$ ) * 0.3 = $V_b$ * (T/60) ……(1)

The relative velocity of the buses in the opposite direction of the cyclist = $V_b$ + $V_c$

Buses go past him in the opposite direction = every 6 m = 6/60 h

Distance covered = ( $V_b$ + $V_c$ ) * 1/10

Since the buses leave every T minute, we get ( $V_b$ + $V_c$ ) * 1/10 = $V_b$ * T/60 …….(2)

From equations (1) & (2), we get

( $V_b$ + $V_c$ ) / 10 = ( $V_b-V_c$ ) * 0.3

$V_b$ + $V_c$ = 3( $V_b$ - 3 $V_c$ )

2 $V_b$ = 4 $V_c$ , $V_b$ = 2 $V_c$ So $V_b$ = 2 * 20 = 40 kmph

From the equation (1), we get

(40-20) * 0.3 = 40 * T/60, T = 9 min