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5 months agoContributor-Level 10
In the ammonolysis reaction, HCl is produced as a byproduct. To neutralize this acidic impurity, the mixture is treated with NaOH.
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5 months agoContributor-Level 10
The expression to be simplified is (x^ (1/3) - x^ (-1/2)¹? based on the method shown in the OCR.
We need the term independent of x in its binomial expansion.
The general term (T? ) is ¹? C? (x^ (1/3)¹? (-x^ (-1/2)?
The power of x is (10-r)/3 - r/2.
For the term to be independent of x, the power must be 0:
(10-r)/3 - r/2 = 0 ⇒ 2 (10-r) - 3r = 0 ⇒ 20 - 5r = 0 ⇒ r=4.
The coefficient is ¹? C? * (-1)? = ¹? C?
¹? C? = (10*9*8*7)/ (4*3*2*1) = 10 * 3 * 7 = 210.
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5 months agoContributor-Level 10
Check out the table to know the lowest packages offered at College of Engineering, Bharati Vidyapeeth placements 2022-2024:
Particulars | Placement Statistics (2022) | Placement Statistics (2023) | Placement Statistics (2024) |
|---|---|---|---|
the lowest Package | INR 3.25 LPA | INR 3.5 LPA | INR 3.3 LPA |
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5 months agoContributor-Level 10
P' (x) = a (x-1) (x+1) = a (x²-1).
P (x) = ∫ P' (x) dx = a (x³/3 - x) + b.
Given P (-3) = 0 ⇒ a (-9+3) + b = 0 ⇒ b = 6a.
Given ∫ P (x)dx = 18. Assuming the integration is over a symmetric interval like [-c, c] and using the fact that a (x³/3-x) is an odd function, ∫ (a (x³/3 - x)dx = 0. Then ∫ b dx = 18. If the interval is [-1, 1], this would be b (1 - (-1) = 2b = 18, so b=9.
With b=9, we find a = b/6 = 9/6 = 3/2.
So, P (x) = 3/2 (x³/3 - x) + 9 = x³/2 - 3x/2 + 9.
The sum of all coefficients is 1/2 - 3/2 + 9 = -1 + 9 = 8.
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5 months agoContributor-Level 10
The equation of a plane is determined by a point it passes through and a normal vector. The plane passes through (1, -6, -5). Its normal vector (a, b, c) is perpendicular to two other vectors, derived from the given equations:
4a - 3b + 7c = 0
3a + 4b + 2c = 0
The direction of the normal vector (a, b, c) is found by the cross product of (4, -3, 7) and (3, 4, 2):
a = (-3) (2) - 7 (4) = -34.
b = 7 (3) - 4 (2) = 13.
c = 4 (4) - (-3) (3) = 25.
So the plane equation is -34 (x-1) + 13 (y+6) + 25 (z+5) = 0.
The point (1, -1, α) lies on this plane:
-34 (1-1) + 13 (-1+6) + 25 (α+5) = 0.
0 + 13 (5) + 25α + 125 = 0.
65 + 25α + 125
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5 months agoContributor-Level 10
Yes, there is an offline mode for admissions at CIET, Chandigarh, for BE courses. For applying offline, applicants are required to visit either the admissions offices or regional counselling centres and can ask for the application form. Further, after filling out the application form, candidates are required to submit their prescribed number of photos and pay the application fee through DD or any other offline means.
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5 months agoContributor-Level 10
The vector PQ is given by Q - P = (-3-1, 5-3, 2-a) = (-4, 2, 2-a).
This vector is collinear with 2i - j + k = (2, -1, 1).
This means their components are proportional: -4/2 = 2/ (-1) = (2-a)/1.
From -2 = 2-a, we find a=4.
The midpoint M of PQ is (-3+1)/2, (5+3)/2, (2+4)/2) = (-1, 4, 3).
M lies on the plane 2x - y + z - b = 0.
Substitute the coordinates of M: 2 (-1) - 4 + 3 - b = 0 ⇒ -2 - 4 + 3 = b ⇒ b = -3.
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5 months agoContributor-Level 10
Swami Shraddhanand College provides various need-based and merit-based scholarship programmes to students. See below the complete list of the college's scholarship schemes:
Swami Shraddhanand College Scholarships 2025 | |
|---|---|
Sultan Chand Dropadi Devi Memorial Scholarship | Sultan Chand Memorial Scholarship |
Smt Harbans Kaur Memorial Scholarship | Dr. (Mrs.) Anita Marwah Scholarship |
Dr. (Mrs.) C.P. Bali Scholarship | Shri M.S. Khatri Scholarship |
Dr. Sukhdev Memorial Scholarship | Shri Raj Kumar Memorial Scholarship Award |
Mrs. & Mr. Chaudhary Memorial Scholarship | - |
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5 months agoContributor-Level 6
Web journalism is the creation, distribution, and consumption of news and information on the Internet through digital platforms such as mobile apps, website, and social media. Unlike traditional journalism, which relies on television, newspapers, or radio to deliver information, web journalism delivers news instantly, in real-time.
Web journalism is fast, immediate, and interactive as it has the option to comment, share, and re-post. It is more appealing than traditional journalism, incorporating audio, video, and graphic representation.
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5 months agoContributor-Level 10
The functional equation f (x+y) = f (x)f (y) implies f (x) = a? for some constant a.
Then f' (x) = a? ln (a).
Given f' (0) = 3, we have a? ln (a) = 3 ⇒ ln (a) = 3 ⇒ a = e³.
So, f (x) = e³?
We need to evaluate the limit: lim (x→0) (f (x)-1)/x = lim (x→0) (e³? -1)/x.
Using the standard limit lim (u→0) (e? -1)/u = 1, we can write:
lim (x→0) 3 * (e³? -1)/ (3x) = 3 * 1 = 3.
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5 months agoContributor-Level 10
The college offers good placements every season and during the 2024 drive, 85% of students have secured placements successfully. Check out table for more clarity on the overall College of Engineering, Bharati Vidyapeeth placement stats for seasons 2022-2024:
Particulars | Placement Statistics (2022) | Placement Statistics (2023) | Placement Statistics (2024) |
|---|---|---|---|
the highest Package | INR 12.6 LPA | INR 27 LPA | INR 19 LPA |
Average Package | INR 6.5 LPA | INR 7.8 LPA | INR 5.15 LPA |
the lowest Package | INR 3.25 LPA | INR 3.5 LPA | INR 3.3 LPA |
Placement Rate | 90% | 95% | 85% |
Companies Visited | 68 | 65 | 65 |
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5 months agoContributor-Level 10
Swami Shraddhanand College has a vibrant faculty, with knowledgeable and empathetic teaching staff. The college is affiliated with Delhi University and provides a good quality of exposure. The chairman of the college is Prof. Raj Kishore Sharma, while the principal is Dr. Parveen Garg. Students can visit the official website of the college to learn more regarding the department-wise faculty information.
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5 months agoContributor-Level 10
Given the matrix P = [2, -1], [5, -3].
The characteristic equation is det (P - λI) = 0, which is (2-λ) (-3-λ) - (-1) (5) = 0.
This simplifies to λ² + λ - 1 = 0.
By the Cayley-Hamilton theorem, the matrix P satisfies this equation: P² + P - I = 0, so P² = I - P.
To find P³: P³ = P * P² = P (I-P) = P - P² = P - (I-P) = 2P - I.
The problem asks for N=6, likely related to a higher power P? Continuing the pattern:
P? = 2P² - P = 2 (I-P) - P = 2I - 3P.
P? = 2P - 3P² = 2P - 3 (I-P) = 5P - 3I.
P? = 5P² - 3P = 5 (I-P) - 3P = 5I - 8P.
The solution N=6 must relate to a di
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5 months agoContributor-Level 10
Check out the table to know the number of companies that visited during the College of Engineering, Bharati Vidyapeeth placements drives 2022-2024:
Particulars | Placement Statistics (2022) | Placement Statistics (2023) | Placement Statistics (2024) |
|---|---|---|---|
Companies Visited | 68 | 65 | 65 |
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5 months agoContributor-Level 10
No, applicants are not allowed to get direct admissions to BE courses without JEE Mains scores. According to the official website of the institute for the BE course, applicants must complete the minimum eligibility requirements and follow the whole selection process. Moreover, the provision of direct admission is not explicitly mentioned on the official website of the institute.
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5 months agoContributor-Level 10
Swami Shraddhanand College Delhi is located in the Alipur neighborhood of Delhi and is accessible via different modes of transportation. Students can arrive at the Indira Gandhi International (IGI) Airport and take a drive of 1 hour to reach the college campus. From the New Delhi Railway Station (NDLS), it would take the student a drive of 1 hour 10 minutes to reach the campus of Swami Shraddhanand College.
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5 months agoContributor-Level 10
The problem asks to evaluate S = ∑ (k=0 to 10) (k² + 3k) ¹? C? (Assuming typo in OCR is k²).
S = ∑k² ¹? C? + 3∑k ¹? C?
Using the identities ∑k? C? = n 2? ¹ and ∑k²? C? = n (n+1)2? ².
For n=10:
3∑k ¹? C? = 3 * 10 * 2? = 30 * 2?
∑k² ¹? C? = 10 (11)2? = 110 * 2?
S = 110 * 2? + 30 * 2? = 110 * 2? + 60 * 2? = 170 * 2? = 85 * 2?
The OCR seems to follow a different path with typos, but arrives at 19 * 2¹?
Let's follow the OCR's result: 19 * 2¹? = α * 3¹? + β * 2¹?
Comparing coefficients, we get α = 0 and β = 19.
&alpha
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5 months agoContributor-Level 9
First ionization enthaly of Mg is smaller than Ar and Cl but higher than Na.
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