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10 months ago

Nehru College Nehru College of Pharmacy

Can I get into Nehru College of Pharmacy with 90 percentile in KEAM?

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saurya snehal

Contributor-Level 10

The Nehru College of Pharmacy accepts KEAM results for admission to various programmes. Following theNehru College of Pharmacy cutoffs from the previous year, the KEAM cutoff percentile for candidates in the General All India category was less than 90. According to the cutoff pattern from previous years, a candidate with a 90 percentile can therefore be admitted to Nehru College of Pharmacy.

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10 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen Maths

By examining the chest X-ray, the probability that TB is detected when a person is actually suffering is 0.99. The probability of a healthy person diagnosed to have TB is 0.001. In a certain city, 1 in 1000 people suffers from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB?

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} : t h e e v e n t t h a t a p e r s o n h a s T B \\ E_{2} : t h e e v e n t t h a t a p e r s o n d o e s n o t h a v e T B \\ L e t H b e t h e e v e n t t h a t t h e p e r s o n i s d i a g n o s e d t o h a v e T B . \\ B a y e s' T h e o r m \\ ∴ P (E_{1}) = \frac{1}{1000} = 0.001, P (E_{2}) = 1 - \frac{1}{1000} = \frac{999}{1000} = 0.999 \\ P (H / E_{1}) = 0.99, P (H / E_{2}) = 0.001 \\ ∴ P (E_{1} / H) = \frac{P (E_{1}) . P (H / E_{1})}{P (E_{1}) . P (H / E_{1}) + P (E_{2}) . P (H / E_{2})} \\ = \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.001} = \frac{0.99}{0.99 + 0.999} \\ = \frac{0.990}{0.990 + 0.999} = \frac{990}{1989} = \frac{110}{221} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{110}{221} . \end{array}$

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10 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen Maths

There are three urns containing: 2 white and 3 black balls 3 white and 2 black balls 4 white and 1 black ball respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn, and it is found to be white. Find the probability that the ball drawn was from the second urn.

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

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10 months ago

Maulana Abul Kalam Azad University of Technology Maulana Abul Kalam Azad University of Technology

What is the MAKAUT NIRF ranking under the Engineering category?

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Vishakha

Contributor-Level 10

Recently, MAKAUT NIRF ranking is going to be announced soon by the NIRF. However, the university has been ranked continuously over the years for its Engineering category. To know the exact ranks allocated by NIRF to B.Tech courses at MAKAUT, refer to the table below:

Publisher	2022	2023	2024
NIRF	201-250	101-150	123

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10 months ago

Kaziranga University Colleges in Jorhat

Is MBA free at Kaziranga University?

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Ashwin Yadav

Contributor-Level 10

Kaziranga University offers the popular MBA programme to students looking for a career in Business Management. The total tuition fees for Kaziranga University fees is INR 4.60 lakhs. Adding additional costs like hostel fees and one-time charges, the total cost that the student would have to bear would be around INR 6.58 lakhs. In addition to this, students can explore the various scholarship options at Kaziranga University.

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10 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen Maths

There are two bags, one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls. A die is thrown. If it shows up 1 or 3, a ball is taken from the first bag; if it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball.

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} b e t h e e v e n t o f s e l e c t i n g B a g I \\ E_{2} b e t h e e v e n t o f s e l e c t i n g B a g I I \\ a n d E_{3} b e t h e e v e n t t h a t b l a c k b a l l i s s e l e c t e d \\ ∴ P (E_{1}) = \frac{2}{6} = \frac{1}{3} a n d P (E_{2}) = 1 - \frac{1}{3} = \frac{2}{3} \\ P (E_{3} / E_{1}) = \frac{3}{7} a n d P (E_{3} / E_{2}) = \frac{4}{7} \\ ∴ P (E_{3}) = P (E_{1}) . P (E_{3} / E_{1}) + P (E_{2}) . P (E_{3} / E_{2}) \\ = \frac{1}{3} . \frac{3}{7} + \frac{2}{3} . \frac{4}{7} = \frac{3 + 8}{21} = \frac{11}{21} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{11}{21} . \end{array}$

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10 months ago

IIIT Nagpur - Indian Institute of Information Technology btech

How many BTech students were placed during the recent IIIT Nagpur placements?

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Shoaib Mehdi

Contributor-Level 10

As per the latest report, the BTech batch 2025 comprised 332 students, of whom 207 registered for the placements and 201 got placed in reputed companies. The total no. of students registered and placed during IIIT Nagpur placements 2024 are presented below:

Particulars	BTech Placement Statistics (2025)
Total students	332
Total Eligible Students	207
Students placed	201

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10 months ago

Jaipuria Indore - Jaipuria Institute of Management Colleges in Indore

What is the LLM fees and placement in Jaipuria Institute of Management, Indore?

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10 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen Maths

A letter is known to have come either from TATA NAGAR or from CALCUTTA. On the envelope, just two consecutive letters "TA" are visible. What is the probability that the letter came from TATA NAGAR?

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} : T h e e v e n t t h a t t h e l e t t e r c o m e s f r o m T A T A N A G A R \\ a n d E_{2} : T h e e v e n t t h a t t h e l e t t e r c o m e s f r o m C A L C U T T A \\ A l s o, E_{3} : T h e e v e n t t h a t o n t h e l e t t e r, t w o l e t t e r s T A a r e v i s i b l e . \\ ∴ P (E_{1}) = \frac{1}{2}, P (E_{2}) = \frac{1}{2} a n d P (\frac{E_{3}}{E_{1}}) = \frac{2}{8} a n d P (\frac{E_{3}}{E_{2}}) = \frac{1}{7} \\ [? F o r T A T A N A G A R, t h e t w o consecutive l e t t e r s v i s i b l e a r e T A, A T, T A, A N, N A, A G, G A, A R] \\ ∴ P (E_{3} / E_{1}) = \frac{2}{8} \\ a n d [F o r C A L C U T T A, t h e t w o consecutive l e t t e r s v i s i b l e a r e C A, A L, L C, C U, U T, T T a n d T A] \\ S o, P (E_{3} / E_{2}) = \frac{1}{7} \\ N o w using B a y e s' T h e r o r m, w e h a v e \\ ∴ P (E_{1} / E_{3}) = \frac{P (E_{1}) . P (E_{3} / E_{1})}{P (E_{1}) . P (E_{3} / E_{1}) + P (E_{2}) . P (E_{3} / E_{2})} \\ = \frac{\frac{1}{2} . \frac{2}{8}}{\frac{1}{2} . \frac{2}{8} + \frac{1}{2} . \frac{1}{7}} = \frac{\frac{1}{8}}{\frac{1}{8} + \frac{1}{14}} = \frac{\frac{1}{8}}{\frac{7 + 4}{56}} = \frac{7}{11} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{7}{11} . \end{array}$

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10 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Four physics ncert solutions class 11th

For two vectors A and B, A B A B + = − is always true when (a) A B = ≠ 0 (b) A⊥ B (c) A B = ≠ 0 and A and B are parallel or anti parallel (d) when either A or B is zero.

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Vishal Baghel

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- |A+B|=|A-B|

$\sqrt[]{|{A |}^{2} + |{B |}^{2} + 2| A| | B | c o s θ} = \sqrt[]{|{A |}^{2} + |{B |}^{2} - 2| A| | B | c o s θ}$

$|{A |}^{2} + |{B |}^{2} + 2| A| | B | c o s θ$ = $|{A |}^{2} + |{B |}^{2} - 2| A| | B | c o s θ$

4|A|B|cos $θ$ =0

|A|²+|B|²cos $θ$ =0

A=0 or B=0 so $θ = 90$ . so A perpendicular B

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10 months ago

Study Abroad Business Management Studies

How is MBA programme in IMD University Switzerland?

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Shreya Basu

Contributor-Level 10

MBA degree from IMD University in Switzerland is considered one of best programs to choose for students willing to study in Switzerland for MBA. Designed specifically for professionals who want to excel in field of leadership, administration and management, university gives outstanding global opportunities to students for hands on experience with real world problems. Program goes on for 12 months. According to unofficial sources, graduates from this program end up with salary hike of 100% after 3 years of graduation. Some top industries where graduates after MBA program are hired include Consulting, Finance, Technology and Consumer goo

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10 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Four physics ncert solutions class 11th

For a particle performing uniform circular motion, choose the correct statement(s) from the following:

(a) Magnitude of particle velocity (speed) remains constant.

(b) Particle velocity remains directed perpendicular to radius vector.

(c) Direction of acceleration keeps changing as particle moves.

(d) Angular momentum is constant in magnitude but direction keeps changing.

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Vishal Baghel

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, c

Explanation- (i) speed will constant throughout

(ii) velocity will be tangential in the direction of motion

(iii) centripetal acceleration will be a= v²/r, will always be towards centre of the circular path.

(iv) angular momentum is constant in magnitude and direction out of the plane perpendicularly as well.

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10 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen Maths

A shopkeeper sells three types of flower seeds: $A_{1}$ , $A_{2}$ , and $A_{3}$ . They are sold as a mixture where the proportions are 4:4:2, respectively. The germination rates of the three types of seeds are 45%, 60%, and 35%. Calculate the probability:

(i) Of a randomly chosen seed to germinate

(ii) That it will not germinate given that the seed is of type $A_{3}$

(iii) That it is of type $A_{2}$ given that a randomly chosen seed does not germinate.

0 Follower 24 Views

alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t A_{1} : A_{2} : A_{3} = 4 : 4 : 2 \\ ∴ P (A_{1}) = \frac{4}{10}, P (A_{2}) = \frac{4}{10} a n d P (A_{3}) = \frac{2}{10} w h e r e A_{1}, A_{2} a n d A_{3} a r e t h e t h r e e t y p e s o f s e e d s . \\ L e t E b e t h e e v e n t s t h a t a s e e d a n d \bar{E} b e t h e e v e n t s t h a t a s e e d d o e s n o t \\ ∴ P (\frac{E}{A_{1}}) = \frac{45}{100}, P (\frac{E}{A_{2}}) = \frac{60}{100} a n d P (\frac{E}{A_{3}}) = \frac{35}{100} \\ a n d P (\frac{\bar{E}}{A_{1}}) = \frac{55}{100}, P (\frac{\bar{E}}{A_{2}}) = \frac{40}{100} a n d P (\frac{\bar{E}}{A_{3}}) = \frac{65}{100} \\ (i) P (E) = P (A_{1}) . P (\frac{E}{A_{1}}) + P (A_{2}) . P (\frac{E}{A_{2}}) + P (A_{3}) . P (\frac{E}{A_{3}}) \\ = \frac{4}{10} . \frac{45}{100} + \frac{4}{10} . \frac{60}{100} + \frac{2}{10} . \frac{35}{100} \\ = \frac{180}{1000} + \frac{240}{1000} + \frac{70}{1000} = \frac{490}{1000} = 0.49 \\ (i i) P (\bar{E} / A_{3}) = 1 - P (E / A_{3}) = 1 - \frac{35}{100} = \frac{65}{100} = 0.65 \\ (i i i), B a y e s' T h e o r e m, w e g e t \\ P (A_{2} / \bar{E}) = \frac{P (A_{2}) . P (\bar{E} / A_{2})}{P (A_{1}) . P (\bar{E} / A_{1}) + P (A_{2}) . P (\bar{E} / A_{2}) + P (A_{3}) . P (\bar{E} / A_{3})} \\ = \frac{\frac{4}{10} . \frac{40}{100}}{\frac{4}{10} . \frac{55}{100} + \frac{4}{10} . \frac{40}{100} + \frac{2}{10} . \frac{65}{100}} = \frac{\frac{160}{1000}}{\frac{220}{1000} + \frac{160}{1000} + \frac{130}{1000}} \\ = \frac{160}{220 + 160 + 130} = \frac{160}{510} = \frac{16}{51} = 0.314 \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{16}{51} o r 0.314 \end{array}$

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10 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Four physics ncert solutions class 11th

Following are four different relations about displacement, velocity and acceleration for the motion of a particle in general. Choose the incorrect one (s) :

(a) V_av= $\frac{1}{2} [v (t_{1}) + v (t_{2})]$

(b) V_av= $\frac{r {(t}_{2}) - r {(t}_{1})}{t_{2} - t_{1}}$

(c) V_av= $\frac{1}{2} [v (t_{1}) - v (t_{2}) (t_{2} - t_{1})]$

$(d) V a v = \frac{v {(t}_{2}) - v {(t}_{1})}{t_{2} - t_{1}}$

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Vishal Baghel

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, c

Explanation – as we know average acceleration is a_av= $\frac{? v}{? t} = \frac{v_{2} - v_{1}}{t_{2} - t_{1}}$

But when acceleration is not uniform V_av is not equal to v₁+v₂/2

So we can write $? v = \frac{? r}{? t}$

$? r = r_{2} - r_{1}$ = v₂-v₁ (t₂-t₁)

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10 months ago

IIIT Nagpur - Indian Institute of Information Technology B.Tech

How good are the BTech placements at IIIT Nagpur?

0 Follower 11 Views

Shoaib Mehdi

Contributor-Level 10

IIIT Nagpur concluded the placements for its BTech batch 2025 with an 97.10% placement rate. The highest package and average package offered to BTech students during IIIT Nagpur placements 2025 stood at INR 60 LPA and INR 14.96 LPA, respectively. Check out the key highlights of IIIT Nagpur placements for the BTech Class of 2025 in the table below:

Particulars	BTech Placement Statistics (2025)
the highest package	INR 60 LPA
Average package	INR 14.96 LPA
Median package	INR 12 LPA
Lowest package	INR 8 LPA
Total students	332
Total Eligible Students	207
Students placed	201
Placement rate	97.10%

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10 months ago

Rajagiri College of Social Sciences Business Management Studies

How can I apply for MBA scholarships at Rajagiri College of Social Sciences?

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Himanshi Pandey

Contributor-Level 10

Rajagiri College of Social Sciences offers two scholarships to MBA students. To apply for these scholarships, candidates must visit the college's official website and submit a duly filled application form before the stipulated deadline. The college's scholarships committee reviews the received applications and grants scholarships to eligible candidates. In addition to fulfilling the course-specific eligibility and selection criteria, candidates must satisfy the following RCSS Kerala MBA scholarships eligiblity criteria:

Diversity Scholarships: Applicant must be a non-Keralite
Career and Talent Scholarships: Applicant must have score

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10 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Four physics ncert solutions class 11th

A particle slides down a frictionless parabolic (y + x² ) track (A – B – C) starting from rest at point A (Fig. 4.2). Point B is at the vertex of parabola and point C is at a height less than that of point A. After C, the particle moves freely in air as a projectile. If the particle reaches highest point at P, then

(a) KE at P = KE at B

(b) Height at P = height at A

(c) Total energy at P = total energy at A

(d) Time of travel from A to B = time of travel from B to P.

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Vishal Baghel

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- c

Explanation– as the given track y=x² is a frictionless track thus total energy will be same throughout the journey.

Hence total energy at A = total energy at P . at B the particle is having only Ke but at P some KE is converted to P

Hence (KE)_B= (KE)_P

Total energy at A = PE= total energy at B = KE= total energy at P

= PE+KE

Potential energy at A is converted to KE and PE at P hence

(PE)P< (PE)A

Hence (height)P= (height)A

As height of p < height of A

Hence path length AB > path length BP

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10 months ago

Dental BDS

I am a Kashmiri RBA girl scoring 362 marks in NEET 2025 and 289120 rank. Can I get a government BDS seat in Kashmir?

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10 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Four physics ncert solutions class 11th

Two particles are projected in air with speed vo at angles θ1 and θ2 (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices

(a) Angle of projection : q1 > q2

(b) Time of flight : T1 > T2

(c) Horizontal range : R1 > R2

(d) Total energy : U1 > U2 .

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Vishal Baghel

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a,b,c

Explanation – H= $\frac{u^{2} {s i n}^{2} θ}{2 g}$

H₁=V_o²sin^{2
$θ$}₁/2g , H₂=V_o²sin^{2
$θ$}₂/2g

H₁>H₂

V_o²sin^{2
$θ$}₁/2g= V_o²sin^{2
$θ$}₂/2g

Sin^{2
$θ$}₁>sin^{2
$θ$}₂

Sin^{2
$θ$}₁ – sin² $θ$ ₂>0

(Sin $θ$ ₁ – sin $θ$ ₂)( Sin $θ$ ₁ + sin $θ$ ₂)>0

Sin $θ$ ₁>sin $θ$ ₂ or ₁ >₂

T= $\frac{2 u s i n θ}{g} = \frac{2 v_{o} s i n θ}{g}$

T₁= $\frac{2 v_{o} s i n ϑ_{1}}{g}$ , T₂= $\frac{2 v_{o} s i n ϑ_{2}}{g}$

T₁> T₂

R= $\frac{u^{2} s i n 2 θ}{g} = \frac{v_{o}^{2} s i n 2 θ}{g}$

Sin $θ$ ₁>sin $θ$ ₂

Sin2 $θ$ ₁> sin2 $θ$ ₂

$\frac{R_{1}}{R_{2}} = \frac{S i n 2 θ_{1}}{s i n 2 θ_{2}} 1$

R₁>R₂

Total energy for the first particle

U₁=K.E+P.E=1/2m_{1
$v_{o}^{2}$}

U₂= K.E+P.E= 1/2m_{2
$v_{o}^{2}$}

Total energy for the second particle

So m₁= m₂ then U₁=U₂

So m₁>m₂ then U₁>U₂

So m₁2 then U1

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10 months ago

Humanities & Social Sciences TUEE Preparation

How can I prepare myself for TUEE 2025 to get into MA in education?

0 Follower 5 Views

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