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

Explanation – T_sand= $\frac{AP+QC}{1}+\frac{PQ}{v}=\frac{25\sqrt[]{2}+25\sqrt[]{2}}{1}+\frac{50\sqrt[]{2}}{v}$

 = 50 $\sqrt[]{2}+\frac{50\sqrt[]{2}}{v}=50\sqrt[]{2}\left(\frac{1}{v}+1\right)$

Time taken T_outside= $\frac{AR+RC}{1}s$

AR= $\sqrt[]{{75}^2+{25}^2}=25\sqrt[]{10}m$

RC= AR= $25\sqrt[]{10}m$

T_outside= 2AR= 50 $\sqrt[]{10}s$

T_sandoutside  . so After solving we get velocity is greater v= 0.81m/s

This is a multiple choice answer as classified in NCERT Exemplar

(c) The pressure inside the gas will be 

P= p_a+Mg/A

A= area of piston

P_a= atmospheric pressure

Mg = weight of piston

When temperature is increases

pV=nRT so volume increases at constant pressure.

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The UPSC IES 2026 Mains exam will be conducted on June 21, 2026. It is a descriptive type exam. That assesses candidates' technical knowledge and problem-solving abilities in engineering disciplines. The Engineering Services Mains Examination  consists of two conventional-type papers in engineering disciplines, each with a duration of three hours and a maximum score of 600 (300 marks in each paper).

Of course, internships and industrial visits are an important part of the MFTech program. Usually, the first three semesters of the course have theoretical subjects, and the last semester is allocated for internships, project work, industrial visits, etc. These activities help students gain exposure to the real working world, polish their communication skills, and grow their network. Hope this will help you :)

As of now, the college has not specified an official deadline for BBA applications for the 2025 academic year. Since admissions generally follow the declaration of Class 12 results, and the application window is currently open, candidates are advised to apply at the earliest. It's best to keep checking the official website for any updates regarding the closing date.

This is a multiple choice answer as classified in NCERT Exemplar

(b) Boyle's law is applicable when temperature is constant

PV=nRT=constant

PV= constant

So pressure is inversely proportional to volume.

So process is called isothermal process.

Yes, admissions for the 2025 BBA programme at Nazareth College of Arts and Science are open. Eligible candidates are encouraged to visit the official website and complete the application process without delay, as seats are limited and deadlines may close soon.

In 2025, Shikshapeeth College of Management and Technology recorded a placement rate above 90%.However, Shikshapeeth College of Management and Technology provides good placement to its students. The data tabulated below provides Shikshapeeth College of Management and Technology placement highlights for 2025:

After NBSAT result declaration, candidates will be shortlisted for the Personal Interview (PI) round. They will have to appear for the PI round on the given dates and submit Statement of Purpose (SOP). Post the PI round, all the candidates will be evaluated on their performance and overall academic profile. Those who shall pass the admission criteria will be offered admission.

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

Explanation – r=cos $\theta ?+sin\theta$ ?…….1

$\theta =-sin\theta ?+cos\theta ?$ …….2

Multiplying eq1 by sin $\theta$  and 2 with cos $\theta$ and adding

Rsin $\theta +\theta cos\theta =sin\theta cos\theta ?+{sin}^2\theta j+{cos}^2\theta ?-sin\theta cos\theta ?$

= ?( ${cos}^2\theta +{sin}^2\theta$ )=j

= rsin $\theta +\theta cos\theta =j$

 n(rcos $\theta -\theta sin\theta$ )=i

b)r $\theta =cos\theta ?+sin\theta ?(-sin\theta ?+cos\theta ?)$

 = -cos $\theta sin\theta +sin\theta .cos\theta =0$ $\theta =90$

c)r=cos $\theta ?+sin\theta ?$

dr/dt=d/dt(cos $\theta ?+sin\theta ?$ )=w[-cos $\theta ?+sin\theta ?$ ]

d)L= M^oLT⁰

e)a=1unit , r= $\theta r=\theta [\mathrm{c}\mathrm{o}\mathrm{s}\theta ?+sin\theta ?]$

v= dr/dt= $\frac{d\theta }{dt}r+\theta \frac{d}{dt}[\mathrm{c}\mathrm{o}\mathrm{s}\theta ?+sin\theta ?]$

v= $\frac{d\theta }{dr}r+\theta [-\mathrm{c}\mathrm{o}\mathrm{s}\theta ?+sin\theta ?]\frac{d\theta }{dt}$

= $\frac{d\theta }{dt}r+\theta w\theta =wr+$ w $\theta \theta$

a= $\frac{d}{dt}\left[wr+w\theta \theta \right]=\frac{d}{dt}\left[\frac{d\theta }{dt}r+\frac{d\theta }{dt}\left(\theta \theta \right)\right]$

a= $\frac{d^2\theta }{d{t}^2}r+\frac{d\theta }{dt}$ $\frac{dr}{dt}+\frac{d^2\theta }{d{t}^2}\theta \theta +\frac{d\theta }{dt}\frac{d}{dt}\left(\theta \theta \right)$

= $\frac{d^2\theta }{d{t}^2}r+w^2\theta +\frac{d^2\theta }{d{t}^2}\theta \theta +w^2\theta +w^2\theta (-r)$

This is a multiple choice answer as classified in NCERT Exemplar

(d) In an ideal gas when a molecules collides elastically with a wall, the momentum transferred to each molecule will be twice the magnitude of its normal momentum. For the face EFGH, it transfer only half of that.

Yes, Shikshapeeth College of Management and Technology provides summer internship opportunities to its students. Check out the table below to know more about the placement rate during Shikshapeeth College of Management and Technology summer internships:

In 2025, Over 50 companies were part of Shikshapeeth College of Management and Technology placement session. Refer to the table below to know more:

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

Explanation- a) for x direction u_x= u+v_ocos $\theta$

u_y=velocity in y direction= v₀sin $\theta$

now tan $\theta =\frac{u_y}{u_x}=\frac{u_{o}sin\theta }{u+u_{o}cos\theta }$

$\theta ={tan}^{-1}\frac{u_{o}sin\theta }{u+u_{o}cos\theta }$

b) let t be the time flight y =0 u_y=v_osin $\theta ,a_y=-g,t=T$

y= u_yt+1/2 a_yt²

0= v_osin $\theta T$ + $\frac{1}{2}\left(-g{T}^2\right)$

So T = $\frac{2u_{o}sin\theta }{g}$

c) horizontal range R, = (u+v_ocos $\theta$ T= (u+v_ocos $\theta$ ) $\frac{2u_{o}sin\theta }{g}$

d) for range to be maximum dR/d $\theta =0$

$\frac{v_o}{g}\left[2ucos\theta +v_{o}cos2\theta \times 2\right]=0$

$2ucos\theta +2v_o\left[2{cos}^2\theta -1\right]=0$

4v_ocos^{2 $\theta +2ucos\theta -2v_o=0$}

So cos $\theta$ = $\frac{-u?\sqrt[]{u^2+8v_o^2}}{4v_o}$

$\theta ={cos}^{-1}\left[\frac{-u\pm \sqrt[]{u^2+8v_o^2}}{4v_o}\right]$

e) cos $\theta$ = $\frac{-v_o?\sqrt[]{u^2+8v_o^2}}{4v_o}=\frac{-1+3}{4}=\frac{1}{2}$

so $\theta =60$

f) if u=0 ${\theta }_{max}={cos}^{-1}\left[\frac{-0\pm \sqrt[]{u^2+8v_o^2}}{4v_o}\right]={cos}^{-1}\left(\frac{1}{\sqrt[]{2}}\right)=45$ ⁰

Yes, it is mandatory to appear for the PI round of NBSAT as it is an essential part of selection process for MBA/PGDM courses. Merely scoring high in NBSAT will not be enough for admission. If a candidate shortlisted for NBSAT PI round does not appear for it, their candidature will be cancelled and the seat will be offered to the next candidate.

This is a multiple choice answer as classified in NCERT Exemplar

(b) As the motion of the vessel as a whole does not affect the relative motion of the gas molecules with respect to the walls of the vessel, hence pressure of the gas inside the vessel, as conserved by us, on the ground remains the same.

If you clear NBSAT exam but do not have the minimum marks in graduation, then you might not get selected for further admission round. To be eligible for admission to MBA/PGDM programme of NBS, meeting the minimum criteria for 60 percent marks in graduation is essential.

Course wise placements for Shikshapeeth College of Management and Technology has not been released yet. However, students can refer to the key highlights of the Shikshapeeth College of Management and Technology placement statistics given in the table below: