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11 months ago

Assertion and Reason:

Directions:

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

(d) Assertion is wrong statement but reason is correct statement.

10.1.Assertion: The elements of Group 2 are commonly known asthe alkaline earth metals.

Reason: Theiroxides and hydroxides are alkaline in nature and thesemetal oxides are found in the earth’s crust.

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11 months ago

Are the graduating students of TMU placed in good companies?

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diksha soni

Contributor-Level 10

The graduating students of TMU are reported to have been placed in good companies across the globe. The top recruiters include Haier, TCS, Infosys, Wipro, etc. The table below provides the placement highlights recorded during TMU placements 2025:

Particulars

Placement Statistics (2025)

the highest Package

INR 63 LPA (overall)

INR 20 LPA (CCSIT)

INR 15 LPA (MBA)

the highest International package

INR 32 LPA

Placement Offers

2500

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11 months ago

Physics Ncert Solutions Class 12th Electrostatic Potential and Capacitance

2.27 A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

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alok kumar singh

Contributor-Level 10

2.27 Capacitance of the charged capacitor, $C_{1}$ = 4 $μ F = 4 \times 10^{- 6}$ F

Supply voltage, $V_{1}$ = 200 V

Electrostatic energy stored in the $C_{1}$ capacitor, $E_{1}$ = $\frac{1}{2} C_{1} {V_{1}}^{2}$ = $\frac{1}{2} \times 4 \times 10^{- 6} \times 200^{2}$ J

= 0.08 J

Capacitance of the uncharged capacitor, $C_{2}$ = 2 $μ F$ = 2 $\times 10^{- 6}$ F

When $C_{2}$ is connected to $C_{1}$ , the potential acquired by $C_{2}$ be $V_{2}$

From the conservation of energy, the charge acquired by $C_{1}$ becomes the charge acquired by $C_{1}$ and $C_{2} .$

Hence $V_{2} \times (C_{1} + C_{2}) = C_{1} V_{1}$

or $V_{2}$ = $\frac{C_{1} V_{1}}{(C_{1} + C_{2})}$ = $\frac{4 \times 10^{- 6} \times 200}{(4 \times 10^{- 6} + 2 \times 10^{- 6})}$ = $\frac{400}{3}$ V

Electrostatic energy of the combination of two capacitors is given by

$E_{2}$ =

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2.27 Capacitance of the charged capacitor, $C_{1}$ = 4 $μ F = 4 \times 10^{- 6}$ F

Supply voltage, $V_{1}$ = 200 V

Electrostatic energy stored in the $C_{1}$ capacitor, $E_{1}$ = $\frac{1}{2} C_{1} {V_{1}}^{2}$ = $\frac{1}{2} \times 4 \times 10^{- 6} \times 200^{2}$ J

= 0.08 J

Capacitance of the uncharged capacitor, $C_{2}$ = 2 $μ F$ = 2 $\times 10^{- 6}$ F

When $C_{2}$ is connected to $C_{1}$ , the potential acquired by $C_{2}$ be $V_{2}$

From the conservation of energy, the charge acquired by $C_{1}$ becomes the charge acquired by $C_{1}$ and $C_{2} .$

Hence $V_{2} \times (C_{1} + C_{2}) = C_{1} V_{1}$

or $V_{2}$ = $\frac{C_{1} V_{1}}{(C_{1} + C_{2})}$ = $\frac{4 \times 10^{- 6} \times 200}{(4 \times 10^{- 6} + 2 \times 10^{- 6})}$ = $\frac{400}{3}$ V

Electrostatic energy of the combination of two capacitors is given by

$E_{2}$ = $\frac{1}{2}$ ( $C_{1} + C_{2}) \times V_{2}^{2}$ = $\frac{1}{2} \times$ ( $4 \times 10^{- 6} +$ 2 $\times 10^{- 6}) \times {(\frac{400}{3})}^{2}$ = 5.33 $\times 10^{- 2}$ J

Hence the amount of electrostatic energy lost by capacitor $C_{1} = E_{1} - E_{2}$

= 0.08 - 5.33 $\times 10^{- 2}$ J = 0.0267 J = 2.67 $\times 10^{- 2}$ J

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11 months ago

The S Block Elements The S-block Elements

Assertion and Reason:

Directions:

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

(d) Assertion is wrong statement but reason is correct statement.

10.1.Assertion: The elements of Group 2 are commonly known asthe alkaline earth metals.

Reason: Theiroxides and hydroxides are alkaline in nature and thesemetal oxides are found in the earth’s crust.

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11 months ago

Teerthanker Mahaveer University Colleges in Moradabad

Does the TMU provide 100% placements?

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diksha soni

Contributor-Level 10

The placement rate for 2025 has not been released. The TMU recorded a placement rate of 85% during the 2024 placement drive. The institute is known for its good placements that has been successful in placing graduating students over the years. Check out the table below for more information on TMU placements for the year 2025:

Particulars

Placement Statistics (2025)

the highest Package

INR 63 LPA (overall)

INR 20 LPA (CCSIT)

INR 15 LPA (MBA)

the highest International package

INR 32 LPA

Placement Offers

2500

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11 months ago

Marri Laxman Reddy Institute of Technology and Management (MLRITM) Marri Laxman Reddy Institute of Technology and Management (MLRITM) Eligibility

What are the criteria for M.Tech at Marri Laxman Reddy Institute of Technology and Management?

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Divya Dixit

Contributor-Level 9

Marri Laxman Reddy Institute of Technology and Management (MLRITM) offers M.Tech with a duration of two years. Candidates seeking admission to M.Tech courses must have a graduation degree.The admission to the M.Tech course is entrance-based. The institute accepts various entrance exams including GATE and PGECET for admission.

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11 months ago

Xavier Law School, Xavier University, Bhubaneswar Colleges in Orissa - Other

Can I get admission into Xavier Law School, XIM University with 75% in Class 12?

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Anupama Garg

Contributor-Level 10

Yes, you may get admission into Xavier Law School, XIM University with 75% in Class 12. However, the institute admits students based on entrance exam scores and personal interviews. Hence, Class 12 marks are not enough to secure a spot at the institute. However, candidate's performance in the last qualifying exam may play a part while deciding the merit list.

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11 months ago

Dr. Shakuntala Misra National Rehabilitation University Colleges in Lucknow

What is the application deadline for BTech admission at DSMNRU Lucknow?

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Anya Vimal

Contributor-Level 10

The Dr. Shakuntala Misra National Rehabilitation University has not released the application deadline for this year yet. The deadline is expected to be announced in July (tentative). Applications for the 2025 academic year are available now for all courses, including the BTech course. Students are seeking to study for a BTech at the university are required to submit a completed application form on the official university website. In order not to miss any deadlines for applications, prospective students are encouraged to visit the official website regularly to obtain updates regarding upcoming events.

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11 months ago

Dr. Shakuntala Misra National Rehabilitation University Accounting & Commerce

What is the application deadline for BCom admission at Dr Shakuntala Misra National Rehabilitation University?

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Nikita Pandey

Contributor-Level 10

The Dr. Shakuntala Misra National Rehabilitation University has not yet released the application deadline. However, the announcement should be in July (tentative). Admission into all courses, including the BCom, is ongoing for the 2025 academic year. Candidates wanting to gain admission into the university's BCom course need to fill out an application form and submit it via the university's official portal. Interested applicants are also advised to check the official website for details on forthcoming events to avoid missing deadlines.

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11 months ago

How are the placements at TMU?

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diksha soni

Contributor-Level 10

The TMU is showing an impressive placement record as of now. Over 400 companies took part in the placement season. Check out the table for more information on TMU placements 2025:

Particulars

Placement Statistics (2025)

the highest Package

INR 63 LPA (overall)

INR 20 LPA (CCSIT)

INR 15 LPA (MBA)

the highest International package

INR 32 LPA

Placement Offers

2500

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11 months ago

Some Basic Concepts of Chemistry Chemistry

1.36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction.

4 HCl (aq) + MnO₂ (s) ———–> 2 H₂O (l) + MnCl₂(aq) +Cl₂(g)

How many grams of HCl react with 5.0 g of manganese dioxide? (Atomic mass of Mn = 55 u

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Payal Gupta

Contributor-Level 10

1.36. The molar mass of MnO₂is 87g and the molar mass of HCl is 36.5 g.

According to the equation, 4 moles of HCl (i.e. 4 x 36.5g = 146 g) reacts with 1 mol of MnO₂ (87g).

So, for 5.0 g of MnO₂ will react with:

= 5 x146/87 = 8.40 g of MnO₂

Therefore, 8.4 g of HCl will react with 5 g of MnO₂.

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11 months ago

Marri Laxman Reddy Institute of Technology and Management (MLRITM) Engineering

How can I get admission to M.Tech at Marri Laxman Reddy Institute of Technology and Management?

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Shiksha Bhandari

Contributor-Level 9

Marri Laxman Reddy Institute of Technology and Management (MLRITM) offers a full-time programme at the PG level. The institute offers 30 seats in the M.Tech programme. Candidates seeking admission to MTech can apply for the admission form online. To be eligible for admission to MTech, the candidates must pass a graduation degree. The college accepts GATE and PGECET for admission to M.Tech courses.

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11 months ago

Some Basic Concepts of Chemistry Chemistry

1.35. Calcium carbonate reacts with aqueous HCl according to the reaction

CaCO₃ (s) + 2HCl (aq) ———->CaCl₂(aq) +CO₂(g) +H₂O(l).

What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?

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Payal Gupta

Contributor-Level 10

1.35. Step 1: 0.75 M HCl means 0.75 mol per 1000 mL or 0.75 x 36.5 g in 1000 mL.

i.e. 1000 mL of 0.75 M HCl contains 0.75 x 36.5 g HCl

Therefore, 25 mL of 0.75 M HCl contains 0.75 x 36.5 x 25/ 1000 g = 0.6844 g

Step 2: To calculate mass of CaCO₃reacting completely with 0.6844 g of HCl
CaCO₃ (s) + 2HC1 (aq)———>CaCl₂ (aq) +CO₂ (g) + H₂O
2 mol of HCl, i.e., 2 x 36.5 g = 73 g HCl react completely with CaC0₃= 1 mol = 100 g

Therefore, 0.6844 g HCl will react completely with CaCO₃= 100/73 x 0.6844 g = 0.938 g

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11 months ago

GLA University Invertis University

What is the Shiksha rating of Invertis University compared to GLA University and Integral University?

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Himanshu Singh

Contributor-Level 10

Invertis University has a Shiksha rating of 3.9 out of 5, reflecting overall student satisfaction with academics and infrastructure. In comparison, GLA University scores slightly higher at 4.1, while Integral University holds a rating of 4.0. These ratings help interested students assess overall reputation and quality before making a decision.

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11 months ago

Some Basic Concepts of Chemistry Chemistry

1.34. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at S.T.P.) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.

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Payal Gupta

Contributor-Level 10

1.34. (i) 1 mole (44 g) of CO₂ will have 12 g carbon.

So, 3.38 g of CO₂ will have carbon = 12g/44g * 3.38

= 0.9217 g

18 g of water will have 2 g of hydrogen.

So, 0.690 g of water contain hydrogen = 2g/18g * 0.6902g

= 0.0767 g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:

= 0.9217 g + 0.0767 g

=0.9984 g

So, the percentage of Carbon in the compound = 0.9217/0.9984 * 100 = 92.32%

Now, percentage of Hydrogen in the compound = 0.0767/0.9984 * 100 = 7.68%

Moles of carbon in the compound = 92.32/12=7.69

Moles of hydrogen in the compound = 7.68/1=7.68

Since, we ha

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11 months ago

Medicine & Health Sciences Pharma

Are there any private B Pharma colleges in Delhi NCR?

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Tasbiya Khan

Contributor-Level 10

Yes, there are many private B Pharma colleges in Delhi NCR. Candidates can see the table below for the top private colleges with tuition fees -

Private Colleges	Tuition Fee
Amity University Noida Admission	INR 9.6 lakh
SVSU Admission	INR 4.3 lakh - INR 4.5 lakh
Galgotias University Admission	INR 5.16 lakh
SGT University Gurgaon Admission	INR 2 lakh - INR 8 lakh
Amity University Gurugram Admission	INR 9.12 lakh
IIMT University Meerut Admission	INR 2.1 lakh - INR 4.2 lakh

Source: Official site and may vary.

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11 months ago

Some Basic Concepts of Chemistry Chemistry

1.33. Calculate the number of atoms in each of the following:

(i) 52 moles of He

(ii) 52 u of He

(iii) 52 g of He

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Payal Gupta

Contributor-Level 10

1.33. (i) 52 moles of Ar

Ans:1 mole of Ar = 6.022 * 10²³ atoms of Ar

Therefore, 52 mole of Ar = 52 * 6.022 * 10²³ atoms of Ar= 3.131 * 10²⁵ atoms of Ar

(ii) 52 u of He

Ans:1 atom of He = 4u of the He

Or, 4 u of He = 1 atom of He

So, 52 u of He = 52/4 atom of He = 13 atoms of He.

(iii) 52 g of He

Ans:4g of He = 6.022 * 10²³ atoms of He

So, 52g of He = 6.022 * 1023 * 52/4 atoms of He

= 7.8286 * 10²⁴ atoms of He

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11 months ago

Physics Ncert Solutions Class 12th Electrostatic Potential and Capacitance

2.26 The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.

(a) How much electrostatic energy is stored by the capacitor?

(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.

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alok kumar singh

Contributor-Level 10

2.26 Area of the parallel plate capacitor, A = 90 ${c m}^{2}$ = 90 $\times 10^{- 4}$ $m^{2}$

Distance between plate, d = 2.5 mm = 2.5 $\times 10^{- 3}$ m

Potential difference across plates, V = 400 V

Capacitance of the capacitor is given by, C = $\frac{ε_{0} A}{d},$

where $ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$ == 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

Electrostatic energy stored in the capacitor is given by the relation

$E_{1}$ = $\frac{1}{2}$ C $V^{2}$ = $\frac{1}{2}$ $\times \frac{ε_{0} A}{d} \times V^{2}$ = $\frac{1}{2}$ $\times \frac{8.854 \times 10^{- 12} \times 90 \times 10^{- 4}}{2.5 \times 10^{- 3}} \times 400^{2}$ = 2.55 $\times 10^{- 6}$ J

Volume of the given capacitor, V’ = A $\times d$ = 90 $\times 10^{- 4} \times$ 2.5 $\times 10^{- 3}$ $m^{3}$

= 2.25 $\times 10^{- 5}$ $m^{3}$

Energy stored is given by u = $\frac{E_{1}}{V ’}$ = $\frac{2.55 \times 10^{- 6}}{2.25 \times 10^{- 5}}$ = 0.113 J/ $m^{3}$

Also, u = $\frac{E_{1}}{V ’}$ = $\frac{\frac{1}{2} C V^{2}}{A d}$ = $\frac{\frac{1}{2} \times \frac{ε_{0} A}{d} \times V^{2}}{A d}$ &

...more

2.26 Area of the parallel plate capacitor, A = 90 ${c m}^{2}$ = 90 $\times 10^{- 4}$ $m^{2}$

Distance between plate, d = 2.5 mm = 2.5 $\times 10^{- 3}$ m

Potential difference across plates, V = 400 V

Capacitance of the capacitor is given by, C = $\frac{ε_{0} A}{d},$

where $ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$ == 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

Electrostatic energy stored in the capacitor is given by the relation

Volume of the given capacitor, V’ = A $\times d$ = 90 $\times 10^{- 4} \times$ 2.5 $\times 10^{- 3}$ $m^{3}$

= 2.25 $\times 10^{- 5}$ $m^{3}$

Energy stored is given by u = $\frac{E_{1}}{V ’}$ = $\frac{2.55 \times 10^{- 6}}{2.25 \times 10^{- 5}}$ = 0.113 J/ $m^{3}$

Also, u = $\frac{E_{1}}{V ’}$ = $\frac{\frac{1}{2} C V^{2}}{A d}$ = $\frac{\frac{1}{2} \times \frac{ε_{0} A}{d} \times V^{2}}{A d}$ = $\frac{1}{2} ε_{0} {(\frac{V}{d})}^{2}$ = $\frac{1}{2} ε_{0} E^{2}$ , where E = $\frac{V}{d}$ electric intensity

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11 months ago

Marri Laxman Reddy Institute of Technology and Management (MLRITM) Marri Laxman Reddy Institute of Technology and Management (MLRITM) Admission Process

What is the seat intake for BTech at Marri Laxman Reddy Institute of Technology and Management?

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Raushan Bhatnagar

Contributor-Level 9

The overall seat intake of BTech courses at Marri Laxman Reddy Institute of Technology and Management is 1710. Candidates must have Class 10 and Class 12 with 60% aggregate from a recognised board for eligible to apply for admission. Further, candidates required to apply and appear for EAMCET entrance exam accepted by the institute. After the fulfillment of the eligibility criteria, candidates have to visit the official website of Marri Laxman Reddy Institute of Technology and Management to apply online.

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11 months ago

Some Basic Concepts of Chemistry Chemistry

1.32. Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes: