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The fee for the MBA programme at NIMS - Nehru Institute of Management Studies is INR 1 lakh. Candidates are required to participate in the admission process, and if they qualify the process they become eligible to pay the course fee and secure their admission to the programme. Candidates must make sure to pay the course fee through the guidelines that have been set by the institute.; 

This is a short answer type question as classified in NCERT Exemplar

As diathermic walls allow exchange of heat energy between two systems and adiabatic walls do not, hence diathermic walls are used to make the bulb of a thermometer.

The NIFTEE Exam for the B.Des programme doesn't ask very difficult maths questions; so, even if you are not from Science background and have average skills in the subject, you can manage. Just solve previous years question and sample papers, and read book for NIFT preparation.

The total fees for the UG course is inclusive of various components. These are either one-time chargeable or have to be paid annually or semester-wise. Sant Baba Bhag Singh University fee for the BBA programme is presented below:

NOTE: The aforementioned fee is as per the data curated from the official website/ sanctioning body. It is still subject to changes and hence, is indicative.

The total course fee for BPharm at S.P. Group of Colleges is INR 1.60 lakh. Candidates have to pay some part of the fee at the time of admission to confirm their seat. Besides, aspirants opting for hostels need to pay a hostel fee of INR 35,000 annually. Aspirants can pay the fee in the mode specified by the college. 

The following documents are needed to complete the NIMS - Nehru Institute of Management Studies application form:

Marksheet of Class 10
Marksheet of Class 12

Marksheet of Graduation
Transfer Certificate

TANCET result

Government-issued ID
Candidates can finish the application and admissions procedure for the courses at NIMS - Nehru Institute of Management Studiesby presenting the aforementioned papers.

This is a long answer type question as classified in NCERT Exemplar

(a) Power radiated by stefan’s law

P= $\sigma A{T}^4=($ 4 $\pi$ R²)T⁴

= 5.67 $\times {10}^{-4}\times 4\times 3.14\times {0.5}^2\times {10}^6\times 4$

= 1.78 $\times {10}^{17}J/s$ =1.8 $\times {10}^{17}J/s$

(b) Energy available per second U= 1.8 $\times {10}^{17}J/s$ = 18 $\times 10$ ¹⁶J/s

Actual energy required to evaporate water = 10%of 18 $\times {10}^{17}j/s$

=1.8 $\times {10}^{16}$ J/s

Energy used per second to raise the temperature of m kg of water from 30⁰C to 100⁰C and then into vapour at 100⁰C

=ms_{w $?\theta$} +mL= m $\times 4186\times \left(100-30\right)+m\times 22.6\times {10}^5$

= 2.93 $\times {10}^{5}m+22.6\times {10}^{5}m=22.53\times {10}^{5}mJ/s$

As per question , 25.53 $\times {10}^{5}m$ = 1.8 $\times {10}^{16}$

m= $\frac{1.8\times {10}^{16}}{25.53\times {10}^5}=7\times {10}^{9}kg$

(c) Momentum per unit time

P= U/c= $\frac{1.8\times {10}^{17}}{3\times {10}^8}=6\times {10}^{8}kg-m/s$ ²

Momentum per unit time

Area p= p/4 $\pi R^2=\frac{6\times {10}^8}{4\times 3.14\times {\left({10}^3\right)}^2}$

d=47.7N/m²

If you want a short and effective answer, then NIFT entrance preparation books by Arihant are considered to be reliable. They touch all the syllabus and are also recommended by toppers. Also, note that the GAT paper for B.Des carries only 30% weightage, so it is not wise to get bogged down by heavy-duty books.

Candidates who have passed Class 12 in the Science stream with Physics, Chemistry, and Biology/ Mathematics subjects with a minimum of 50% aggregate are eligible to apply for BPharm at S.P. Group of Colleges. Aspirants must hold the documents proving their eligibility for verification at the time of admission.

This is a long answer type question as classified in NCERT Exemplar

Consider the diagram

Θ= θ₁+ θ₂/2

Let temperature varies linearly in the rod from its one end to other end, let θ be the temperature of the midpoint of the rod. At steady state

Rate of flow of heat,

dQ/dt = $\frac{KA({\theta }_1-\mathrm{\theta })}{\frac{L_o}{2}}=\frac{KA\left(\mathrm{\theta }-{\mathrm{\theta }\mathrm{}}_2\right)}{\frac{L_o}{2}}$

where k is the coefficient of thermal conductivity of the rod

so θ₁- θ= θ- θ₂

θ= θ₁+ θ₂/2

L=L₀ (1+ $\alpha \mathrm{\theta }$ )

L= L_o (1+ $\frac{\alpha \left({\mathrm{\theta }\mathrm{}}_1+{\mathrm{\theta }}_2\right)}{2}$ )

Instructions are played at the beginning of the test along with a sample question. After that, you can read section 1 questions, and listen to section 1 and answer them.

This is a long answer type question as classified in NCERT Exemplar

By applying Pythagoras theorem in given figure

^{${\left(\frac{L+?L}{2}\right)}^2={\left(\frac{L}{2}\right)}^2+x$ 2}

x= $\sqrt[]{{\left(\frac{L+?L}{2}\right)}^2-{\left(\frac{L}{2}\right)}^2}$

=1/2 $\sqrt[]{{(L+?L)}^2-L^2}$

= ½ $\sqrt[]{L^2+{?L}^2+2L?L-L^2}$

=1/2 $\sqrt[]{\left({?L}^2+2L?L\right)}$

As Increase in $?L$  is very small so we neglect it

=1/2 $\times \sqrt[]{2?L}$

$?L=L\alpha ?t$

By using this value in above in equation

x=1/2 $\times \sqrt[]{2L\times L\alpha ?T}$ = ½ L $\sqrt[]{2\alpha ?t}$

= $\frac{10}{2}\times \sqrt[]{2\times 1.2\times {10}^{-5}\times 20}$

= 5 $\times \sqrt[]{4\times 1.2\times {10}^{-4}}$

= 5 $\times 2\times 1.1\times {10}^{-2}=0.11=11cm$

Yes, if you have 60% in PCB and English, you can take admission to BSc Nursing at American NRI College of Nursing. All applications should be submitted online and you have to wait for the merit list. After being shortlisted, you need to finish the counseling and submit payment to confirm your status as an admitted student. You can visit the park without buying an entrance ticket.

After getting the necessary marks in PCB and English in Class 12, you may apply for the course. Direct admission to American NRI College of Nursing is given to students depending on their merit. People must be 17 years old to apply. Visit the website and submit your application.

S.P. Group of Colleges offers scholarship amount ranging between INR 16,000 and INR 2.84 lakh. The scholarship amount varies depending on the course applied by the candidate. Below is the course-wise scholarship amount:

This is a long answer type question as classified in NCERT Exemplar

Decrease in temperature $?t$ = 57-37= 20⁰C

Coefficient of linear expansion $\alpha$ = 1.7 $\times {10}^{-5}/$ ^oC

Bulk modulus for copper B = 140 $\times {10}^{9}N/m^2$

Coefficient of cubical expansion $\gamma$ = 3 $\alpha$ = 5.1 $\times {10}^{-5}/°C$

Let initial volume of the cavity be V and its volume increases by $?V$  due to increase in temperature.

$?V=\gamma V?t$

$\frac{?V}{V}=\gamma ?t$

Thermal stress produced = B $\times volumetricstrain$

= B $\times \frac{?V}{V}=B\gamma ?t$

= 140 $\times {10}^9\times 5.1\times {10}^{-5}\times 20$

= 1428 $\times {10}^{8}N/m$ ²

There is no free education at American NRI College of Nursing. In addition, students could try looking for scholarships coming from other sources. You can obtain all the information about fees on the official website. Students have to make payment for admission and tuition, if required by the university.