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11 months ago

Sets Maths

8. Which of the following sets are finite or infinite

(i) The set of months of a year

(ii) {1, 2, 3, . . .}

(iii) {1, 2, 3, . . ., 99, 100}

(iv) The set of positive integers greater than 100

(v) The set of prime numbers less than 99

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Payal Gupta

Contributor-Level 10

8. (i) The set of months of a year has 12 elements. Here, the set is finite.

(ii) The given set has the natural number as its elements. Hence, the set is infinite.

(iii) The given set has 100 elements i.e., from 1 to 100. Hence, the set is finite.

(iv) There are infinite numbers of positive integers greater than 100. Hence the set is infinite.

(v) The numbers of prime number less than 99 is finite. Hence, the set is finite.

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11 months ago

Business Management Studies BBA

What is the placement statistics for the BBA at Gokul Global University?

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Anushka Bidhi

Contributor-Level 10

As per the official reports of Gokul Global University placement, the students get an average salary package of INR 7 LPA. The specific reports for the BBA course is however not available. The rate of placement for the previous years was marked at 78.16% with 1052 students being placed.

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11 months ago

Nursing Delhi

Is it worth joining the top BSc Nursing colleges in Delhi NCR?

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Tasbiya Khan

Contributor-Level 10

Yes, joining the top BSc Nursing colleges in Delhi NCR can be worthwhile for you. For that, you need to check its ROI. Listed below are some colleges along with their tuition fees and average placement package:

College Name	Total Tuition Fee	Average Placement Package
AIIMS Delhi Placement	INR 2,400	INR 24 LPA
Ahilya Bai College of Nursing DU Placement	INR 1,00	INR 8.5 LPA
GGSIPU Placement	INR 3.72 lakh	INR 8.15 LPA
Amrita School of Nursing Faridabad Placement	INR 8 lakh	INR 7.24 LPA
Amrita Vishwa Vidyapeetham Placement	INR 8 lakh	INR 7.24 LPA

Disclaimer: This information is sourced from official website/ media reports/ NIRF website and may vary.

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11 months ago

Teaching & Education KTET

What is the benefit of KTET Exam?

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Satyendra Shukla

Contributor-Level 6

The benefit of clearing the KTET exam is that it makes candidates eligible to apply for teacher recruitment in the Kerala schools. Candidates can apply for Lower Primary/ Upper Primary/ High School teachers posts after clearing the exam.

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11 months ago

Nursing Nursing in Delhi/NCR

Are there any affordable BSc Nursing colleges in Delhi NCR?

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Tasbiya Khan

Contributor-Level 10

Yes, there are many affordable BSc Nursing colleges in Delhi NCR. Some of them are mentioned below along with their tuition fees:

College Name	Tuition Fee	Eligibility / Exams
Lady Hardinge Medical College, University of Delhi Fees	INR 29,420	NEET
Pt. Bhagwat Dayal Sharma Post Graduate Institute of Medical Sciences Fees	INR 48,000	Merit-Based
All India Institute of Medical Sciences Delhi Fees	INR 2,400	AIIMS Paramedical
Raj Kumari Amrit Kaur College of Nursing Fees	INR 1,000	NEET and Merit-Based
Ahilya Bai College of Nursing DU Fees	INR 1,000	NEET

Disclaimer: This information is sourced from official website and may vary.

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11 months ago

Manchester University University of Leeds

Which university is better: University of Manchester or University of Leeds?

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Rachit Sharma

Contributor-Level 10

The University of Manchester and University of Leeds are some of the top universities in the UK. Both the universities offer high placement rate, top-notch infrastructure, impressive ROI, best degree programs and more for its international students. Check the comparison listed below between University of Manchester and University of Leeds:

Parameters	University of Manchester	University of Leeds
Establishment	2004	1904
Location	Manchester, Greater Manchester, UK	Leeds, West Yorkshire, UK
Shiksha Popularity Rankings	#27 in World	#68 in World
Acceptance Rate	39%	64%
Placement Rate	Over 90%	90%
Average Salary	INR 27.74 LPA	INR 26.26 LPA
QS Ranking 2025	#34	#82
UG Annual Fee	INR 18.78 L - 19.1 L	INR 55.34 L - 60.83 L
PG Annual Fee	INR 20.57 L - 47.23 L	INR 16.96 L - 98.87 L
Popular Streams	Mechanical Engineering \| Chemical Engineering \| Physics	Computer Science \| Aerospace Engineering \|Arts and Design

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11 months ago

Business Management Studies BBA

Can I get direct admission in the BBA course at Gokul Global University?

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Anushka Bidhi

Contributor-Level 10

Usually there is no provision for student to get direct admission in the BBA at Gokul Global University. The students are selected based of the result of Class 12. Additionally the students are required to satisfy the set eligibility criteria by the students.

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11 months ago

Teaching & Education KTET Pattern

What is the pass mark for KTET 2025?

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Rupali Pruthi

Contributor-Level 10

The pass mark for KTET 2025 exam differ for reserved and unreserved category candidates. Have a look at the KTET passing marks below:

For General and OBC: 60 per cent marks (90 marks out of 150)
For SC/ ST: 55 per cent marks (82 marks out of 150)

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11 months ago

NIPS Hotel Management, Bhubaneswar Business Management Studies

Is admission open for the MBA at NIPS Hotel Management Bhubaneswar?

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Himanshu Singh

Contributor-Level 10

Yes, admission is currently open for the MBA at NIPS Hotel Management Bhubaneswar. Interested candidates are encouraged to apply at the earliest and regularly check the official website for updates on deadlines, entrance exam details, and other admission-related announcements.

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11 months ago

B.Sc Career

What are the career prospects after BSc from PTSNS University?

0 Follower 4 Views

Aishwarya Kumari

Contributor-Level 10

PTSNS University BSc graduates can find job opportunities in both government and public sectors. The job profile and salary package can vary based on the specialisation in which the course is done. A lot of students also opt for higher studies. Some of the government departments, where the degree holders can find opportunities are Forest Department, Indian Railways, AIIMS, Indian Agriculture Research Institute, Food Corporation of India, etc. Read more about Career after BSc course.

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New Question

11 months ago

Sets Maths

Class 11 Chapter 1 Set Ex 1.2 NCERT Solutions

7. Which of the following are examples of the null set

(i) Set of odd natural numbers divisible by 2

(ii) Set of even prime numbers

(iii) {x : x is a natural numbers, x < 5 and x > 7}

(iv) {y : y is a point common to any two parallel lines}

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Payal Gupta

Contributor-Level 10

1. (i) There is no odd number that can be divided by 2.

?The given set is a null set.

(ii) An even prime number is 2. Hence, the set will have 2 as element.

? The given set is not a null set.

(iii) As there is no natural number which is both less than 5 and greater than 7.

? The given set is a null set.

(iv) Two parallel lines never meets and hence no common point.

? The given set is a null set.

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11 months ago

Amar Nath Girls Degree College

How can I register online for admission at Amar Nath Girls Degree College?

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Lakshita Saini

Beginner-Level 5

To register online for admission at Amar Nath Girls Degree College:

Go to the official website of the college.
Click on the “Admission” or “Apply Online” option.
Select the course you want to apply for.
Fill in your personal and educational details in the form.
Upload scanned copies of your documents like photo, mark sheets, and certificates.
Pay the application fee online if required.
Submit the form and note your application number for future reference.

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11 months ago

Exam Study Abroad

Can I get a full scholarship with SAT?

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Nishtha Sharma

Contributor-Level 6

Full scholarships can be difficult solely based on SAT scores. But applicants can avail more than 25% of the scholarship amount. But applicants will also be required to have a good academic background and if not, they must have exemplary SOPs or admission essays along with LORs.

Applicants must have SAT scores above average like 1050 in 2023, in order to also aim for scholarships.

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New Question

11 months ago

Sets Maths

Class 11 Chapter 1 Set Ex 1.2 NCERT Solutions

Q1. Which of the following are examples of the null set

(i) Set of odd natural numbers divisible by 2

(ii) Set of even prime numbers

(iii) {x : x is a natural numbers, x < 5 and x > 7}

(iv) {y : y is a point common to any two parallel lines}

0 Follower 4 Views

New Question

11 months ago

Teaching & Education KTET

What is the KTET exam for?

0 Follower 31 Views

Rashmi Shekhar

Contributor-Level 6

The KTET is a state-level exam conducted every year to evaluate the quality of candidates who aspire to become teachers in Lower Primary, Upper Primary and High Schools in Kerala. The Kerala Pareeksha Bhavan conducts the Kerala TET exam twice in a year to determine the eligibility of candidates as teachers. Those who clear the exam become eligible to apply for teacher recruitment exams in the state schools.

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11 months ago

physics ncert solutions class 11th System of Particles and Rotational Motion

7.23 A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m².

(a) What is his new angular speed? (Neglect friction)

(b) Is kinetic energy conserved in the process? If not, from where does the change come about?

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Vishal Baghel

Contributor-Level 10

(a)

Moment of Inertia of the man-platform system = 7.6 kg-m²

Moment of inertia when the man stretches his hands to a distance 90 cm

= 2 x mr² = 2 x 5 x (0.9)2 = 8.1 kg-m²

Initial moment of inertia of the system, $-$ = 7.6 + 8.1 = 15.7 kg-m²

Angular speed $-$ = 30 rev/min

Angular momentum, $-$ = $I_{i}$ = 15.7 x 30 …… (i)

Moment of inertia when the man folds his hands to a distance of 20 cm

= 2 x mmr²= 2 x 5 x (0.2)2 = 0.4 kg-m²

Final moment of inertia, $ω_{i}$ = 7.6 + 0.4 = 8 kg-m²

Final angular speed = $L_{i}$ and final angular of momentum, $I_{i} ω_{i}$ = $I_{f}$ = 8 $ω_{f}$ …… (ii)

From t

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(a)

Moment of Inertia of the man-platform system = 7.6 kg-m²

Moment of inertia when the man stretches his hands to a distance 90 cm

= 2 x mr² = 2 x 5 x (0.9)2 = 8.1 kg-m²

Initial moment of inertia of the system, $-$ = 7.6 + 8.1 = 15.7 kg-m²

Angular speed $-$ = 30 rev/min

Angular momentum, $-$ = $I_{i}$ = 15.7 x 30 …… (i)

Moment of inertia when the man folds his hands to a distance of 20 cm

= 2 x mmr²= 2 x 5 x (0.2)2 = 0.4 kg-m²

Final moment of inertia, $ω_{i}$ = 7.6 + 0.4 = 8 kg-m²

Final angular speed = $L_{i}$ and final angular of momentum, $I_{i} ω_{i}$ = $I_{f}$ = 8 $ω_{f}$ …… (ii)

From the conservation of angular momentum, we have $L_{f}$ = $I_{f} ω_{f}$

$ω_{f}$ = $I_{i} ω_{i}$ = 58.88 rev/min

(b) Kinetic energy is not conserved in the given process. In fact, with the decrease in the moment of inertia, kinetic energy increases. The additional kinetic energy comes from the work done by the man to fold his hands towards himself.

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11 months ago

B.Sc Pandit S. N. Shukla University

How are the placements for BSc at Pandit SN Shukla University?

0 Follower 4 Views

Virajita Mehra

Contributor-Level 10

A concise BSc placements report is not available as of now. However, as per verifies student reviews on Shiksha, the highest and the lowest packages are INR 13 LPA and INR 4 LPA, respectively. In addition, a total of 400 students were placed during the PTSNS University placements drive in 2023. The given information is not as per official sources and hence, is indicative.

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11 months ago

physics ncert solutions class 11th System of Particles and Rotational Motion

7.22 As shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 m/s2)

(Hint: Consider the equilibrium of each side of the ladder separately.)

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Vishal Baghel

Contributor-Level 10

The given situation can be shown as

NB = force exerted on the ladder by the floor point B

T = Tension in the rope

BA = CA = 1.6 m, DE = 0.5 m and BF = 1.2 m

Mass of the weight, m = 40 kg

The perpendicular drawn from point A on the floor BC, this intersects DE at mid-point H.

Δ ABI and ΔACI are congruent. Therefore BI = IC, I is the mid-point of BC. DE is parallel to BC.

BC = 2 x DE = 1 m and AF = BA – BF = 0.4 m…… (i)

D is the mid-point of AB, hence we can write

AD = (1/2) x BA = 0.8 ……. (ii)

Using equation (i) and (ii), we get FE = 0.4 m

Hence, F i

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The given situation can be shown as

NB = force exerted on the ladder by the floor point B

T = Tension in the rope

BA = CA = 1.6 m, DE = 0.5 m and BF = 1.2 m

Mass of the weight, m = 40 kg

The perpendicular drawn from point A on the floor BC, this intersects DE at mid-point H.

Δ ABI and ΔACI are congruent. Therefore BI = IC, I is the mid-point of BC. DE is parallel to BC.

BC = 2 x DE = 1 m and AF = BA – BF = 0.4 m…… (i)

D is the mid-point of AB, hence we can write

AD = (1/2) x BA = 0.8 ……. (ii)

Using equation (i) and (ii), we get FE = 0.4 m

Hence, F is the mid-point of AD.

Hence G will also be the midpoint of AH.

ΔAFG and ΔADH are similar

FG = (1/2)DH = (1/2) X 0.25 = 0.125 M

In ΔADH, AH = $\frac{3 A B}{4 g \sin ? θ})$ - $\sqrt \frac{11.46}{19.6}$ ) = $\frac{F G}{D H} = \frac{A F}{A D} = \frac{F G}{D F} = \frac{0.4}{0.8} = \frac{1}{2}$ – $\sqrt ({A D}^{2}$ ) = 0.76 m

For translational equilibrium of the ladder, the upward force should be equal to the downward force, NC + NB = mg = 392 …… (iii)

For rotational equilibrium of the ladder, the net moment about A is

${D H}^{2}$ NB x BI +mg x FG + NC x CI + T x AG – T x AG = 0

$\sqrt {({0.8}^{2}$ NB x 0.5 +40 x g + 0.125 x NC x 0.5 = 0

(NC - NB) x 0.5 = 49

NC - NB = 98 ………. (iv)

Adding equation (iii) and (iv) we get

NC = 245 N, NB = 147 N

For rotational equilibrium of the side AB, considering the moment about A

${0.25}^{2}$ NB x BI + mg x FG + T x AG = 0

$-$ 245 x 0.5 + 40 x 9.8 + T x AG = 0

T = 96.7 N

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11 months ago

Sets Maths

6. Match each of the set on the left in the roster form with the same set on the right described in set-builder form:

(i) {1, 2, 3, 6} (a) {x : x is a prime number and a divisor of 6}

(ii) {2, 3} (b) {x : x is an odd natural number less than 10}

(iii) {M,A,T,H,E,I,C,S} (c) {x : x is natural number and divisor of 6}

(iv) {1, 3, 5, 7, 9} &nbs

...more

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Payal Gupta

Contributor-Level 10

6. (i) (c)

(ii) (a)

(iii) (d)

(iv) (b)

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11 months ago

physics ncert solutions class 11th System of Particles and Rotational Motion

7.21 A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.

(a) How far will the cylinder go up the plane?

(b) How long will it take to return to the bottom?

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Vishal Baghel

Contributor-Level 10

Initial velocity of the cylinder, v = 5 m/s

Angle of inclination, $ω$ = 30 $\sqrt (\frac{2}{3})$

Height reached by the cylinder = h

Energy of the cylinder at point A:

KE rot + KEtrans

(1/2) I $ω$ + (1/2) m $θ$

Energy of the cylinder at point B = mgh

Using the law of conservation of energy

(1/2) I $°$ + (1/2) m $ω^{2}$ = mgh

Moment of inertia of the solid cylinder I = (1/2) mr²

Hence (1/2)(1/2)mr² $v^{2}$ + (1/2) m $ω^{2}$ = mgh

We also know v = r $v^{2}$

(1/4)m $ω^{2}$ + (1/2) m $v^{2}$ = mgh

(3/4) $ω$ = gh

h = (3/4)( $v^{2}$ = (3/4)(25/9.81) = 1.91 m

In Δ ABC, $v^{2}$ = $v^{2}$ , AB = BC / $v^{2} / g)$ = 3.82 m

Henc

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Initial velocity of the cylinder, v = 5 m/s

Angle of inclination, $ω$ = 30 $\sqrt (\frac{2}{3})$

Height reached by the cylinder = h

Energy of the cylinder at point A:

KE rot + KEtrans

(1/2) I $ω$ + (1/2) m $θ$

Energy of the cylinder at point B = mgh

Using the law of conservation of energy

(1/2) I $°$ + (1/2) m $ω^{2}$ = mgh

Moment of inertia of the solid cylinder I = (1/2) mr²

Hence (1/2)(1/2)mr² $v^{2}$ + (1/2) m $ω^{2}$ = mgh

We also know v = r $v^{2}$

(1/4)m $ω^{2}$ + (1/2) m $v^{2}$ = mgh

(3/4) $ω$ = gh

h = (3/4)( $v^{2}$ = (3/4)(25/9.81) = 1.91 m

In Δ ABC, $v^{2}$ = $v^{2}$ , AB = BC / $v^{2} / g)$ = 3.82 m

Hence the cylinder will travel 3.82 m on the inclined plane

For radius of gyration K, the velocity of the cylinder at the instance when it rolls back to bottom is given by the relation

v = $\sin ? θ$ = $\frac{B C}{A B}$ )

for solid cylinder, $\sin ? 30 °$ = $\sqrt (\frac{2 g h}{1 + \frac{k^{2}}{R^{2}}})$

v = $\sqrt (\frac{2 g A B \sin ? θ}{1 + \frac{k^{2}}{R^{2}}}$ ) = $k^{2}$ g AB $\frac{R^{2}}{2}$ )

The time taken to return to the bottom is :

t = $\sqrt (\frac{2 g A B \sin ? θ}{1 + \frac{1}{2}}$ = AB/ $\sqrt (\frac{4}{3}$ ( $\sin ? θ$ = $\frac{A B}{v}$ ( $\sqrt$ = $\frac{4}{3} g A B \sin ? θ)$ = 0.764 s