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Lingaya's Vidyapeeth has released the placement report for the Class of 2025. The key highlights of are tabulated below:

No, it is not mandatory to update or provide all GATE scores during the COAP registration. Candidates must choose their best valid GATE scores to secure a seat in the participating colleges. The COAP registrations has started and the last date to register for Round I is May 20 (9:00 am). Candidates must visit the website to complete the registration as soon as possible. The GATE COAP Round II registration is scheduled from May 23 to May 26. 

Greater Noida Institute of Technology IPU offers admission to courses based on the entrance exam scores of the candidates. The institute accepts JEE Main score for admission to BTech programme. The institute also offers admission through Management Quota. The institute provides admission to candidates taking admission under the Management Quota based on JEE Main score. 

Hotel Management offers significant room for innovation and creativity, and this is becoming increasingly valued by top recruiters. 

• Hotels are finding new ways to enhance guest experiences.
• Innovation in Culinary Arts, Room Design & Guest Experience, Event Planning & Banqueting, Digital Marketing & Branding, and Front Office & Customer Relations, areas are always in demand. 
• With increasing environmental awareness, hotels are looking for innovative ways to be sustainable.
• Implementing new technologies to improve efficiency and guest satisfaction.

There are botany offical cutoff released by college but in most colleges it is 50% to 60% is the eligibility criteria.

Hope for best

Frequency of string A, $f_A$ = 324 Hz, let the frequency of string B, be = $f_B$

Beat’s frequency, n = 6 Hz

Beat’s frequency is given as n = $\left|f_A\pm f_B\right|$ or 6 = 324 $\pm f_B$

$f_B=318Hzor330Hz$

Frequency decreases with a decrease in the tension in a string. This is because frequency is directly proportional to the square root of tension, . It is given as

$\nu \propto \surd T$

Hence the beat frequency cannot be 330 Hz. So $f_B=318Hz$

The NID B.Des Provisional Letter for B.Des programme will be released on June 10, 2025. Provisional Offer Letters for B.Des Admissions 2025 will be issued to candidates who have paid the non-refundable token fee and whose uploaded documents have been successfully verified and found to be in order.

Length of the pipe, l = 20 cm = 0.2 m

Source frequency = $n^{th}$ normal mode frequency is given by the relation

${\nu }_n$ = (2n-1) $\frac{\nu }{4l}$ , n is an integer = 0,1,2,3, ….

430 = (2n-1) $\times \frac{340}{4\times 0.2}$

n = 1

Hence, the first mode of vibration frequency is resonantly excited by the given source. In a pipe open at both ends, the $n^{th}$ mode of vibration frequency is given by the relation:

${\nu }_n=\frac{nv}{2l}$ , n = $\frac{2l{\nu }_n}{v}$ = $\frac{2\times 0.2\times 430}{340}$ = 0.5

Since the number of the node of vibration (n) has to be an integer, the given source does not produce a resound vibration in an open pipe.

Yes, PTSNS University BA students can apply for scholarship schemes. The scholarships are provided through the Madhya Pradesh scholarship portal. Additionally, the MP scholarship portal awards scholarships to meritorious students through the following schemes:

• Gaon Ki Beti Yojna
• Mukhya Mantri Medhavi Vidyarthi
• Mukhya Mantri Jankalyan (Shiksha Protsahan Yojna)
• Post Matric Scholarship Scheme (OBC, SC, ST Students)
• Awas Sahayta Yojana (SC, ST Students)
• Pratibha Kiran Scholarship Yojna
• Vikramaditya Scholarship Yojna

Yes, Lingaya's Vidyapeeth conducts annual on campus placement drives for courses like BTech, BBA, BCA, BPharma and more. Here are the highlights:

Length of the steel rod, l = 100 cm = 1 m

Fundamental frequency of vibration,  $\nu$ = 2.53 kHz = 2.53 $\times {10}^3$ Hz

When the rod is plucked at its middle, an antinode (A) is formed at the centre, and the nodes (N) are formed at its two ends, as shown in the figure.

The distance between two successive nodes is $\lambda$ $/2$

l = $\lambda /2$ $\lambda$ = 2l = 2 m

The speed of sound in steel is given by the relation

v = $\nu \lambda$ = 2.53 $\times {10}^3\times 2$ = 5.06 $\times {10}^3$ m/s = 5.06 km/s

Frequency of the tuning fork,  $\nu$ = 340 Hz

Since the given pipe is attached with a piston at one end, it will behave as a pipe with one end closed and the other end open, as shown in this figure.

$l_1$ = $\frac{\lambda }{4},\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}$ of the pipe $l_1$ = 25.5 cm = 0.255 m

$\lambda =4l_1$ = 4 $\times 0.255$ m = 1.02 m

The speed of sound is given by the relation, v = $\nu \lambda$ = 340 $\times 1.02$ = 346.8 m/s

Mass of the wire, m = 3.5 $\times {10}^{-2}$ kg

Linear mass density,  $\mu =\frac{m}{l}$ = 4 $\times {10}^{-2}$ kg/m

Frequency of vibration,  $\nu$ = 45 Hz

Length of the wire, l = $\frac{m}{\mu }$ = $\frac{3.5\mathrm{}\times {10}^{-2}}{4\mathrm{}\times {10}^{-2}}$ = 0.875 m

The wavelength of the stationary wave $\lambda$ = $\frac{2l}{n}$ , where n = number of nodes in the wire

For fundamental node, n = 1,  $\lambda$ = 2l = 2 $\times 0.875=1.75m$

(a) The speed of the transverse wave in the string is given as v = $\nu \lambda$ = 45 $\times 1.75$ = 78.75 m/s

(b) The tension produced in the string is given by the relation, T = $v^2\mu$ = ${\left(78.75\right)}^2$ $\times$ 4 $\times {10}^{-2}$ = 248.06 N

Yes, a candiate can change his/her college preference multiple times before the deadline (May 19, 2025, 5 PM). After the deadline, the last college selected will be considered final.

Yes, SK University does offer BTech courses to its Class 12 passed-out students. This full-time programme is available in various specialisations, such as Mechanical Engineering, Computer Engineering, Civil Engineering, Electrical Engineering, Computer Science and Engineering, etc. For more details, candidates can refer to the following key highlights:

Note: The above mentioned seat intake has been sourced from the official website. It is still subject to change and, hence, is indicative.

(a) The given equation represents a stationary wave because the harmonic terms kx and $\omega t$ appear separately in the equation.

(b) The given equation does not contain any harmonic term. Therefore, it does not represent either a travelling wave or a stationary wave.

(c) The given equation represents a travelling wave as the harmonic terms kx and $\omega t$ are in the combination of kx – $\omega t$ .

(d) The given equation represents a stationary wave because the harmonic terms kx and $\omega t\mathrm{a}\mathrm{p}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{s}\mathrm{e}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{l}\mathrm{y}$ in the equation. This equation represents the superposition of two stationary waves.