Ask & Answer: India's Largest Education Community

Let $\theta$ be the angle made by the radius vector joining the bead and the centre of the wire with the downward direction. Let, N be the normal reaction.

mg = N $\cos ?\theta$ …….(1)

mr ${\omega }^2$ = N $\sin ?\theta$ ……(2)

m(R $\sin ?\theta$ ) ${\omega }^2$ = N $\sin ?\theta$

Hence N = m(R) ${\omega }^2$

Substituting the value on N in eqn (1)

mg = mR ${\omega }^2\cos ?\theta$

or $\cos ?\theta$ = g/ R ${\omega }^2$ ………(3)

As $\left|\cos ?\theta \right|$ $\le$ 1, the bead will remain at the lowermost point

g/ R ${\omega }^2$ $\le 1or\omega \le \sqrt{g}/R$

For $\omega$ = $\frac{\sqrt{2g}}{R},equation\left(3\right)$ becomes

$\cos ?\theta$ = g/ R ${\omega }^2$

$\cos ?\theta$ =(g/R)(R/2g) = ½

$\theta =60°$

Yes, there are approximately 640+ Entrepreneurship colleges in India where students can study. Of these, 420+ are privately owned, 60+ are government-run, and 2 are public-private. Admission to the best Entrepreneurship colleges in India is based on the scores of entrance exams like CAT, MAT, CMAT, XAT, ATMA, MAH CET, etc.

Mass of the man, m = 70 kg

Radius of the drum, r = 3 m

Coefficient of friction between the wall and his clothing,  $\mu$ = 0.15

Number of revs of hollow cylindrical drum = 200 rev/min = 200/60 rev/s = 3.33 rev/s

The centripetal force required is provided by the normal N of the wall on the man

N = m $v^2/R$ = m ${\omega }^2$ R

When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downwards is balanced by the frictional force acting vertically upwards.

The man will not fall, if

mg $\le \mu N$

mg $\le \mu (m{\omega }^{2}R$ )

${\omega }^2\ge g/R\mu$

$g/R\mu$ = 10/ (3 $\times$ 0.15)

$\omega =4.71rad/s$

Yes, Nilamber-Pitamber University (NPU) offers direct admission to the BA programme based on merit. Students who have completed their Class 12 from a recognised board with a minimum aggregate of 45% are eligible to apply. Admissions are primarily merit-based, considering the marks obtained in the qualifying examination. Additionally, NPU accepts CUET UG scores for BA admissions, providing an alternative pathway for applicants. 

When the motorcyclist is at the uppermost point of the death well, the normal reaction R on the motorcyclist by the ceiling of the chamber acts downwards. His weight mg also acts downwards. The outward centrifugal force acting on the motorcyclist is balanced by two forces.

R + mg = m $v^2/r$ , where v is the velocity and m is the combined mass of the motorcycle and motorcyclist

Because of the balance between the forces, the motorcyclist does not fall.

The minimum speed required at the uppermost position to perform a vertical loop is given by R = 0 in the above equation.

So mg = m $v^2/r$ or v = $\sqrt{gr}$ = $\sqrt{10\times 25}$ = 15.8 m/s

To be eligible for the BA course at NPU Jharkhand, candidates must have completed Class 12 from a recognised board such as CBSE or JAC. General category students must secure at least 45% marks in their Class 12 examination. Meeting the minimum eligibility criteria is mandatory for consideration in the admission process.

Speed of revolution of the disc, n = $33\frac{1}{3}$ rev/min = 100/3 rpm = 100/ (3 $\times 60)=0.56rps$

Angular acceleration $\omega$ = 2 $\pi n$ = 2 $\times \frac{22}{7}\times 0.56$ = 3.492 rad/s

The coins revolve with the disc. The centripetal force is provided by the frictional force $m{v}^2/r\le \mu mg$ …. (1)

As v = r $\omega$ , the above equation becomes mr ${\omega }^2/r\le \mu mg$

r $\le \mu g/{\omega }^2$

r $\le$  (0.15 $\times 10)/{3.492}^2$ = 12 cm

For coin A, r = 4 cm

The condition (r $\le$ 12 ) is satisfied for the coin placed at r = 4 cm, so coin A will revolve with the disc.

The condition (r $\le$ 12 ) is not satisfied for the coin placed at r = 14 cm, so coin B will not revolve with the disc.

Yes, Nilamber-Pitamber University (NPU), Jharkhand offers a Bachelor of Arts (BA) programme. It is a three-year undergraduate course structured across six semesters. The university provides this programme across its affiliated colleges, making it accessible for students across the region who are interested in pursuing humanities and social Science disciplines.

To understand this, Assume you have a bucket that has infinite number of apples and if your mother asks "give me the apple". How will you figure out which one is "The Apple", she is asking for.

Similarly any inversre trigonometric functions behaves like a Many-one Function; which means,

For Example $\sin^ {-1}\left (\frac {2} {3}\right)$ can have many solutions, we need to fix one solutions which can be used as standerd value for the function.

• A standerd value (Angles) of any inverse trigonometric value lies between a fiexed range is known as principal value. 

For Ex; The value of $\sin^ {-1}\left (\frac {2} {3}\right)$ will always lie between –? /2 to? /2.

$y={\sin }^{?1}\left(\frac{2}{3}\right)?\sin \left(y\right)=\frac{2}{3}?y?[?\frac{?}{2},\frac{?}{2}]$  

Force on the box, F = MA = 40 $*$ 2 N = 80 N

Frictional force, Ff = $?mg$ = 0.15 $*40*10=$ 60 N

Net force = F – Ff = 80 – 60 = 20 N

From the equation F = ma, we get the backward acceleration produced in the box

a = 20/40 = 0.5 m/s2

From the equation s = ut + $\frac{1}{2}a{t}^2$ , to travel s = 5 m by the box to fall off from the truck, we get

5 = 0 $*t$ + $\frac{1}{2}$ $*$ 0.5 $*t^2$

5 = 0.25 $t^2$ , t = 4.47 s

The travel of truck during t = 4.47 s is

= 0 $*t$ + $\frac{1}{2}$ $*$ 0.5 $*t^2$ = 0.5 $*2*{4.47}^2$ = 19.98 m

Mass of the block = 15 kg

Coefficient of static friction between the block and the trolley  $\mu$ = 0.18

Acceleration of the trolley = 0.5 m/s2

(a) Force experienced by block, F = MA = 15 $\times$ 0.5 = 7.5 N. This fore acts in the direction of motion of the trolley

Force of friction, Ff = $\mu mg$ = 0.18 $\times 15\times 10$ N = 27 N

Force experienced by the block is less than the friction, hence for a stationary observer on the ground, the block will be stationary

(b) When an observer moves with the trolley, the trolley will appear to be at rest

Yes, the Thiagarajar College of Engineering BE/BTech courses are NIRF ranked. In the year 2024, the institute was ranked #147 by the NIRF Rankings under the 'Engineering' category. Earlier, in the year 2023, the institute was ranked #101-150 under the same category by NIRF. Check below to know the Thiagarajar College of Engineering BTech rankings by NIRF:

Asian School of Business offers admission to various UG and PG level courses such as BCom, BCA, BBA and MBA. Among these courses, this business school offers MBA as the flagship course for the duration 2 years. The school offers admission on the basis of merit and previously attempted entrance exam scores. The school considers entrance exams such as CAT, CMAT, KMAT Kerala or other equivalent exams. 

Candidates looking for course admission at CII Institute of Hospitality - Hyatt Kochi must appear for the aptitude test. The selection criteria for this course is based on an aptitude test and an industry panel viva. To get admitted to the CII Institute of Hospitality - Hyatt Kochi, candidates must meet the eligibility criteria set by the institute. Students can also pursue Swiss Professional Degree/Degree in Tourism and Hospitality/Degree in Tourism Studies + Swiss Professional Diploma.

No, the institute admits students based on institute-level aptitude test. The college offers Diploma and Degree programmes in Hotel Management. CII Institute of Hospitality - Hyatt Kochi admissions are offered to students based on their past academic merit of Class 12 and scores of CII Aptitude Test and Industry Panel Viva rounds.

Mass of the block = 25 kg

Mass of the man = 50 kg

Acceleration due to gravity = 10 m/s2

Weight of the block = 250 N

Weight of the man = 500 N

In the 1st case, the man lifts the block directly, applying an upward force, same as the block = 250 N

Due to Newton's 3rd law of motion, the downward reaction of the man on the floor = 500 N + 250 N = 750 N

In the 2nd case, the man applies a downward force of 25 kg wt. According to Newton's 3rd law, the action on the floor by the man 500N-250N = 250N

The man should adopt the second case

CII Institute of Hospitality - Hyatt Kochi admissions for all courses are currently ongoing. Interested candidates can apply in online mode through the official website. The online application process is explained in the following steps:

Step 1: Visit the official website of CII Institute of Hospitality - Hyatt Kochi.

Step 2: Fill out the form.

Step 3: Complete the form and submit it.