Ask & Answer: India's Largest Education Community

Given

Radius, r = 30m

Speed, v = 54 km/h = 15 m/s

The mass of the train, m = 106 kg

(a) The required centripetal force is provided by the rails to the wheels of the train

(b) The angle of banking required to prevent wearing out of the rails

$\tan ?\theta$ = $v^2$ / rg = ${15}^2$ / (30 $\times 10)$

$\theta$ = 37 $°$

Yes, applications are open at CII Institute of Hospitality - Hyatt Kochi for various courses. The mode of application is online. Aspirants need to apply at the official website for admission to the various programme. Candidates can fill out the application form on the institute's official website.

Dr. C.V. Raman University Bihar offers the scholarships to MBA courses for twelve months in the first year. Further, it provides scholarship for 10 months in the second year of the course. Moreover, to continue the scholarship in the second year of the MBA course, candidates are required to achieve a minimum aggregate of 60% in their last qualifying examination. 

The speed of the aircraft, v = 720 km/h = 200 m/s

The angle of banking = 15 $°$

From the relation $\tan ?\theta$ = $v^2$ /rg we get r = $v^2$ / ( g $\times \tan ?\theta )$ = 14928 m

Yes, the exam centre of CUET is available in the Bilaspur, Chattisgarh. It is in between the narmada nagar, sarkanda and Tarbahar chowk. You can also search it on the map and then you will find the exact location.

Asian School of Business Trivandrum offers MBA as a flagship programme for 2 years with an annual intake of 120 students. The school offers respective course in affiliation with CUSAT and with AICTE approval. ASB Trivandrum offers MBA on the basis of merit and valid entrance scores. Candidates must meet the cutoff criteria and get shortlisted for group discussion and personal interview. The remaining shortlisted candidates must pay the fee and confirm the seat.

Speed of water, v = 15 m/s

Cross-sectional area of the tube, A = 10^-2 m²

Density of water $\rho$ = 1000 kg/m3

Mass of the water hitting the wall = $\rho \times A\times v$ = 1000 $\times$ 10^-2 $\times 15$ = 150 kg/s

Force exerted on the wall because of impact of water = mass $\times$ velocity = 150 $\times 15$ N = 2250 N

The net force at the lowest point is denoted by (mg – T₁)and the net force at the highest point is denoted by (mg + T₂ ), hence option (a) is correct. The forces mg and T₁ are in mutually opposite direction at the lowest point and mg and T₂ are in the same direction at the highest point.

BTech in Information Technology is one of the popular courses offered in the field of Technology. After completing this course offered by Thiagarajar College of Engineering, BTech graduates have a plethora of job options. BTech students can apply for jobs across various industries, including Computer Science, Information Technology, AI, and more. Check out the below table for some of the job profiles popular amongst IT graduates to pursue:

The acceleration of the conveyer belt, a = 1 m/s²

Coefficient of static friction,  ${\mu }_s$ = 0.2

Mass of the man, m = 65 kg

The net force experienced by the man, MA = 1 $\times 65$ N = 65 N

This net force is due to the friction between the belt and the man

At maximum static friction

${\mu }_s\times mg=m{a}_{max}$

$a_{max}$ = ${\mu }_s\times g$ = 0.2 $\times 10=2$ m/s²

This graph could be of a ball rebounding between two walls separated by a distance of 2 cm.

The ball rebounds every 2 secs between the walls with uniform velocity.

Velocity of rebounding = displacement / time = ( 2 $*{10}^{-2}$ )/2 = 0.01 m/s

Initial momentum = mu = 0.04 $*0.01=$ 4 $*{10}^{-4}$ kgm/s

Final momentum = -mu = -4 $*{10}^{-4}$ kgm/s

Magnitude of Impulse = Initial momentum – final momentum = 8 $*{10}^{-4}$ kgm/s

The time between two consecutive impulses is 2 secs, so the ball receives an impulse every 2 seconds.

St Xavier's University Kolkata providing admission to MBA specialisation must need to score a percentile of 85 for the General AI category candidates. Further, if a candidate is applying with a OBC category, then he/she must need to score a similar percentile which is 85 to get into the university.

(a) & (c) solution not possible as the stone will fly off tangentially. The answer is (b)

Mass of the stone, m = 0.25 kg

Radius of the circular path = 1.5 m

Speed = 40 rev/min = 0.67 rev/s

Angular velocity,  $?$ = 2 $?n$ = 2 $*\frac{22}{7}*0.67$ = 4.19 rev/s

The tension of the string, T = m ${?}^2$ r = 0.25 $*{4.19}^2*1.5$ = 6.59 N

Maximum tension of the string,  $T_{max}$ = 200 N

From the equation $T_{max}$ = ${mv}_{max}^2$ /r,  $v_{max}^2$ = $T_{max}*r/m$

$v_{max}^2$ = 200 $*1.5$ /0.25

$v_{max}$ = 34.64 m/s