arthham

Exam

ICSE Class 12

Conducting Body
Council for the Indian School Certificate Examinations (CISCE)
Who It Is For
Students completing Class 12
Subjects
Compulsory English and elective subjects from various streams
Question Types
Mix of objective and subjective (short and long answer) questions
Paper Duration
Typically 2 to 3 hours per paper
Marking
Based on final written exams and internal school assessments
Frequency
Held once a year
Qualifies You For
Admission to undergraduate degree courses

The Indian School Certificate (ISC) is the final school-leaving examination for students in Class 12, conducted by the Council for the Indian School Certificate Examinations (CISCE). It is taken by students in schools affiliated with the CISCE board, and the resulting qualification is used for admission to undergraduate programs at universities in India and abroad.

The examination consists of a series of written papers covering a broad curriculum. English is a compulsory subject, and students select several elective subjects from streams such as Science, Commerce, or Humanities. The papers are designed to assess a student's analytical skills and deep understanding of concepts, rather than just rote memorisation.

What makes the ISC examination challenging is the sheer depth and breadth of the syllabus for each subject. The questions demand detailed and well-structured answers, requiring students to apply theoretical knowledge to practical scenarios. Balancing the preparation for multiple in-depth subjects, along with completing practicals and internal assessments, requires consistent effort and effective time management throughout the final two years of school.

How a question gets worked here

From a physics paper rather than ICSE Class 12, worked the way every question on this site is. ICSE Class 12 questions are being added.

JEE Advanced 2012

A small mass mm is attached to a massless string whose other end is fixed at PP as shown in the figure. The mass is undergoing circular motion in the x−yx-y plane with centre at OO and constant angular speed ω\omega. If the angular momentum of the system, calculated about OO and PP are denoted by LOL_O and LPL_P respectively, then
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Step 1: Define the coordinate system and motion parameters.

Let the center of the circle OO be the origin (0,0,0)(0,0,0), and let the fixed point PP lie on the zz-axis at a height hh above OO.

The position vector of the mass mm relative to OO at any time tt is given by:

r⃗O=Rcos⁡θ i^+Rsin⁡θ j^\vec{r}_O = R\cos\theta\,\hat{i} + R\sin\theta\,\hat{j}

where θ=ωt\theta = \omega t.

The linear momentum of the mass is:

p⃗=mv⃗=mωR(−sin⁡θ i^+cos⁡θ j^)\vec{p} = m\vec{v} = m\omega R(-\sin\theta\,\hat{i} + \cos\theta\,\hat{j})

Step 2: Calculate the angular momentum about OO.

The angular momentum about the center OO is defined by the cross product of the position vector and the momentum.

L⃗O=r⃗O×p⃗\vec{L}_O = \vec{r}_O \times \vec{p} L⃗O=(Rcos⁡θ i^+Rsin⁡θ j^)×mωR(−sin⁡θ i^+cos⁡θ j^)\vec{L}_O = (R\cos\theta\,\hat{i} + R\sin\theta\,\hat{j}) \times m\omega R(-\sin\theta\,\hat{i} + \cos\theta\,\hat{j}) L⃗O=mωR2k^\vec{L}_O = m\omega R^2 \hat{k}

Since mm, ω\omega, and RR are constant, L⃗O\vec{L}_O does not change with time.

Step 3: Calculate the angular momentum about PP.

The position vector of the mass relative to PP includes the vertical displacement −hk^-h\hat{k}.

r⃗P=Rcos⁡θ i^+Rsin⁡θ j^−hk^\vec{r}_P = R\cos\theta\,\hat{i} + R\sin\theta\,\hat{j} - h\hat{k}

The angular momentum about PP is:

L⃗P=r⃗P×p⃗\vec{L}_P = \vec{r}_P \times \vec{p} L⃗P=(Rcos⁡θ i^+Rsin⁡θ j^−hk^)×mωR(−sin⁡θ i^+cos⁡θ j^)\vec{L}_P = (R\cos\theta\,\hat{i} + R\sin\theta\,\hat{j} - h\hat{k}) \times m\omega R(-\sin\theta\,\hat{i} + \cos\theta\,\hat{j}) L⃗P=mωR(hcos⁡θ i^+hsin⁡θ j^+Rk^)\vec{L}_P = m\omega R \left( h\cos\theta\,\hat{i} + h\sin\theta\,\hat{j} + R\hat{k} \right)

Step 4: Analyze the time-dependence of both angular momenta.

Express the angular momentum about PP in terms of time tt using θ=ωt\theta = \omega t.

L⃗P=mωRhcos⁡(ωt) i^+mωRhsin⁡(ωt) j^+mωR2k^\vec{L}_P = m\omega Rh\cos(\omega t)\,\hat{i} + m\omega Rh\sin(\omega t)\,\hat{j} + m\omega R^2 \hat{k}

The horizontal component of L⃗P\vec{L}_P rotates in the xx-yy plane, meaning its direction changes continuously with time.

Answer: L⃗O\vec{L}_O remains constant, while L⃗P\vec{L}_P varies with time.

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