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Q. Consider the equation: H=xpϵqErtsH = \frac{x^p \epsilon^q E^r}{t^s}H=tsxpϵqEr​ Where: H = magnetic field E = electric field ε = permittivity x = distance t = time Find the values of p, q, r and s respectively.

Consider the equation   H = (x p ε q E r ) / t s   Where:   H = magnetic field,   E = electric field,   ε = permittivity,   x = distance,   t = time   Find the values of p, q, r and s.   Step 1: Dimensions of Each Quantity Quantity Dimensions H   (Magnetic field) [M T⁻² A⁻¹] E   (Electric field) [M L T⁻³ A⁻¹] ε   (Permittivity) [M⁻¹ L⁻³ T⁴ A²] x   (Distance) [L] t   (Time) [T]   Step 2: Dimensional Equation Substituting dimensions into the equation:   [M T⁻² A⁻¹] = [L]ᵖ × [M⁻¹L⁻³T⁴A²]ᶠ × [MLT⁻³A⁻¹]ʳ × [T]⁻ˢ   Expanding: [M 1 T -2 A -1 ] = [M -q+r L p-3q+r T 4q-3r-s A 2q-r ]   Step 3: Compare Powers on Both Sides Element Equation Result M -q + r = 1 Equation (1) ...

Q. A new unit (α) of length is chosen such that it is equal to the speed of light in vacuum. What is the distance between Venus and Earth in terms of α units, if light takes 6 min 40 s to cover this distance?

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Q. A new unit (α) of length is chosen such that it is equal to the speed of light in vacuum. What is the distance between Venus and Earth in terms of α units, if light takes 6 min 40 s to cover this distance?   Given: New unit α = speed of light in vacuum = 3 × 10⁸ m/s Time for light to travel from Venus to Earth = 6 min 40 s Step 1: Convert time to seconds t = 6  min  40  s = ( 6 × 60 ) + 40 = 360 + 40 = 400  s t = 6 \text{ min } 40 \text{ s} = (6 \times 60) + 40 = 360 + 40 = 400 \text{ s} t = 6 min 40 s = ( 6 × 60 ) + 40 = 360 + 40 = 400 s Step 2: Calculate the actual distance Distance = Speed of light × Time \text{Distance} = \text{Speed of light} \times \text{Time} Distance = Speed of light × Time d = ( 3 × 10 8  m/s ) × 400  s d = (3 \times 10^8 \text{ m/s}) \times 400 \text{ s} d = ( 3 × 1 0 8 m/s ) × 400 s d = 1.2 × 10 11  m d = 1.2 \times 10^{11} \text{ m} d = 1.2 × 1 0 11 m Step 3: Convert distance into α units Since 1 α...

UP Police Computer Operator Free Mock Test | UPPRPB Computer Operator Practice Set

  UP Police Computer Operator Mock Test UP POL UP Police — Computer Operator Mock Test Uttar Pradesh Police Recruitment Board  |  30 Questions  |  90 Marks Time Left 90:00 Answered: 0 Flagged: 0 Unanswered: 30 Q 1 / 30 Computer Fundamentals ⚑ Mark for Review ← Previous Clear Next → Submit Exam Question Map Answered Flagged for Review Not Attempted Current Exam Info Total Questions: 30 Total Marks: 90 Each Question: 3 marks Negative Marking: −1 per wrong Duration: 90 minutes ...

UP Police Computer Operator Free Mock test

UPPRB Computer Operator Mock Test Question 1 | Section: Computer Science Correct: +3.0 Negative: -1.0 Loading question... Mark for Review & Next Clear Response ...

India’s Fuel Crisis: Geopolitics and the Surge in LPG Rationing

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Fueling the Struggle: A street-level view of India's current LPG crisis, capturing the tension between public priority, industrial reduction, and the enforcement of strict new rationing laws.  India is currently navigating a significant commercial LPG shortage triggered by geopolitical instability in the Middle East , specifically the disruption of supply routes through the Strait of Hormuz . This crisis has severely impacted the hospitality sector, leading many hotels and restaurants to appeal to the government for urgent intervention as they face potential closures. In response, the Indian government has prioritised domestic gas supplies to prevent public unrest while implementing strict rationing measures , including a new 25-day mandatory waiting period between residential cylinder bookings. Authorities have also invoked the Essential Commodities Act to stop the illegal stockpiling and black marketing of cylinders during this period of scarcity. To manage the limited invent...

The Digital Panopticon: Technical Analysis of the 149 Million Record Infostealer Repository and the Collapse of Endpoint Trust

  The Digital Panopticon: Technical Analysis of the 149 Million Record Infostealer Repository and the Collapse of Endpoint Trust The dawn of 2026 has been marked by a watershed moment in the landscape of cyber espionage and data exfiltration, characterized by a fundamental shift from platform-centric breaches to the mass industrialization of endpoint harvesting. In January 2026, cybersecurity researcher Jeremiah Fowler identified a sprawling, unprotected, and unencrypted database containing a staggering 149,404,754 unique login and password combinations. 1 This repository, which occupied approximately 96 gigabytes of raw data, represented a centralized collection point for "stealer logs"—the output of malicious software designed to bypass local security controls and siphon digital identities directly from the user's computing environment. 1 Unlike traditional data leaks that originate from the compromise of a central server, such as a social media company or a financial...